1. If $ a,b,c \ge 0$ then
\[ (a \plus{} b \plus{} c)^6 \ge 1728abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\]
2. If $ a,b,c \ge 1/2,abc \equal{} 1$ then
\[ \frac {1}{a \plus{} 2b} \plus{} \frac {1}{b \plus{} 2c} \plus{} \frac {1}{c \plus{} 2a} \le 1
\]
3. If $ a,b,c \ge 0$ then
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \ge \frac {3}{2} \plus{} \frac {27}{16}\cdot \frac {(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}
\]
Hope you like them :)

[quote="can_hang2007"]Problem. Prove the inequality for $ x,y,z \in \mathbb{R}$
\[ \sum \frac {x^2 \minus{} yz}{4x^2 \plus{} 4y^2 \plus{} z^2} \ge 0
\]
My solution.
By expanding, the inequality is equivalent to
\[ 4\sum x^6 \minus{} 4\sum x^5y \minus{} 16\sum xy^5 \plus{} 33\sum x^4y^2 \plus{} 24\sum x^2y^4 \minus{} 20 \sum x^3y^3 \minus{} 4\sum x^4yz \minus{} 20\sum x^3y^2z \minus{} 17\sum x^3yz^2 \plus{} 60x^2y^2z^2 \ge 0
\]
Now, setting $ x \equal{} 4a \plus{} 2b \plus{} c,y \equal{} 4b \plus{} 2c \plus{} a,z \equal{} 4c \plus{} 2a \plus{} b$, the inequality becomes
\[ 22\sum a^4(b^2 \plus{} c^2) \plus{} 32\sum a^3b^3 \minus{} 16\sum a^4bc \minus{} 22 \sum a^2b^2c(a \plus{} b) \minus{} 48a^2b^2c^2 \ge \minus{} 3(\sum a^4b^2 \minus{} \sum a^3bc^2)
\]
By the AM-GM inequality, we have  $ \minus{} 3(\sum a^4b^2 \minus{} \sum a^3bc^2) \le 0$, hence it suffices to prove
\[ 22\sum a^4(b^2 \plus{} c^2) \plus{} 32\sum a^3b^3 \minus{} 16\sum a^4bc \minus{} 22 \sum a^2b^2c(a \plus{} b) \minus{} 48a^2b^2c^2 \ge 0
\]
or
\[ 11\sum a^4(b^2 \plus{} c^2) \plus{} 16\sum a^3b^3 \minus{} 8\sum a^4bc \minus{} 11 \sum a^2b^2c(a \plus{} b) \minus{} 24a^2b^2c^2 \ge 0
\]
Using the assumption $ a \plus{} b \plus{} c \equal{} 1,ab \plus{} bc \plus{} ca \equal{} q,abc \equal{} r$, the last inequality is equivalent to
\[ 3r(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \ge 0
\]
But it's trivial since
\[ 3r(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \ge q^2(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \equal{} q^2(1 \minus{} 3q) \ge 0
\]
We have done. :)[/quote]
Very interesting your solution, can_hang 2007. Your substitution reduces the equality case $ (x,y,z) \equal{} (1,2,4)$ to $ (a,b,c) \equal{} (0,0,1)$. Nice, but very laborious.  :lol:  :o

5. For any $ a,b,c>0,abc\equal{}1$. Prove that inequality
\[ \sum\frac{1}{(a^2\minus{}a\plus{}1)^2} <\frac{11}{3}
\]
:)

Prove that for any $ a,b,c,d,k>0$.
\[ \sum \frac{a^k}{(a\plus{}b\plus{}c)^k} \ge \min (1,\frac{1}{2^{k\minus{}1}},\frac{4}{3^k})
\]
:)

For any $ a,b,c \ge 0$
\[ \sum \sqrt{\frac{a^3}{a^3\plus{}7abc\plus{}b^3}} \ge 1
\]
:)

For any $ a,b,c \ge \frac{1}{2},abc\equal{}1.$ prove that
\[ \sum \frac{a}{b^2\plus{}2c} \ge 1
\]
\[ \sum \frac{a}{b^2\plus{}3c} \ge \frac{3}{4}
\]
:)

For $ a,b,c \ge 0$ then
\[ \sum \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \ge 2\plus{}\frac{1}{\sqrt[3]{2}}
\]
:)

[hide="to tanpham90"] I would like to see your solution :), (without using computer for this inequality). :)[/hide]

For $ a,b,c \ge 0$ then
\[ \sum \sqrt{\frac{8a^2\plus{}bc}{b^2\plus{}c^2}} \ge \frac{9}{\sqrt{2}}
\]
:)

[quote="can_hang2007"]For $ a,b,c \ge 0$ then
\[ \sum \sqrt {\frac {8a^2 \plus{} bc}{b^2 \plus{} c^2}} \ge \frac {9}{\sqrt {2}}
\]
:)[/quote]

I'm looking for a proof without using computer for this inequality. :)

[quote="shfdfzhjj"][quote="can_hang2007"]For $ a,b,c \ge 0$ then
\[ \sum \sqrt [3]{\frac {a^2 \plus{} bc}{b^2 \plus{} c^2}} \ge 2 \plus{} \frac {1}{\sqrt [3]{2}}
\]
:)

[hide="to tanpham90"] I would like to see your solution :), (without using computer for this inequality). :)[/hide][/quote]

Could your give me some hints？
I think all inequalities you have posted are very hard and nice.
I wonder if there exisit nice solution.[/quote]
[hide="hint"] If $ a \ge b \ge c$ then
\[ \frac{a^2\plus{}bc}{b^2\plus{}c^2} \plus{}\frac{b^2\plus{}ca}{c^2\plus{}a^2} \plus{}\frac{6(a^2\plus{}bc)}{a^2\plus{}c^2} \ge \frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}
\]
:)
[/hide]

[quote="can_hang2007"]For any $ a,b,c > 0$
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge 3 \plus{} \frac {6(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}
\]
:)[/quote]
 find k  max such that $ a,b,c>0$  we have  this ineq always right
S.O.S easy if k=4 ,but k=6 maybe hard

[hide="hint"] If $ a \ge b \ge c$ then
\[ \frac {a^2 \plus{} bc}{b^2 \plus{} c^2} \plus{} \frac {b^2 \plus{} ca}{c^2 \plus{} a^2} \plus{} \frac {6(a^2 \plus{} bc)}{a^2 \plus{} c^2} \ge \frac {4(a^2 \plus{} b^2)}{c^2 \plus{} ab}
\]
:)
[/hide][/quote]

But I think that's the way how you create it，it is too hard for me. :(

[quote="can_hang2007"]Problem. Prove the inequality for $ x,y,z \in \mathbb{R}$
\[ \sum \frac {x^2 \minus{} yz}{4x^2 \plus{} 4y^2 \plus{} z^2} \ge 0
\]
My solution.
By expanding, the inequality is equivalent to
\[ 4\sum x^6 \minus{} 4\sum x^5y \minus{} 16\sum xy^5 \plus{} 33\sum x^4y^2 \plus{} 24\sum x^2y^4 \minus{} 20 \sum x^3y^3 \minus{} 4\sum x^4yz \minus{} 20\sum x^3y^2z \minus{} 17\sum x^3yz^2 \plus{} 60x^2y^2z^2 \ge 0
\]
Now, setting $ x \equal{} 4a \plus{} 2b \plus{} c,y \equal{} 4b \plus{} 2c \plus{} a,z \equal{} 4c \plus{} 2a \plus{} b$, the inequality becomes
\[ 22\sum a^4(b^2 \plus{} c^2) \plus{} 32\sum a^3b^3 \minus{} 16\sum a^4bc \minus{} 22 \sum a^2b^2c(a \plus{} b) \minus{} 48a^2b^2c^2 \ge \minus{} 3(\sum a^4b^2 \minus{} \sum a^3bc^2)
\]
By the AM-GM inequality, we have  $ \minus{} 3(\sum a^4b^2 \minus{} \sum a^3bc^2) \le 0$, hence it suffices to prove
\[ 22\sum a^4(b^2 \plus{} c^2) \plus{} 32\sum a^3b^3 \minus{} 16\sum a^4bc \minus{} 22 \sum a^2b^2c(a \plus{} b) \minus{} 48a^2b^2c^2 \ge 0
\]
or
\[ 11\sum a^4(b^2 \plus{} c^2) \plus{} 16\sum a^3b^3 \minus{} 8\sum a^4bc \minus{} 11 \sum a^2b^2c(a \plus{} b) \minus{} 24a^2b^2c^2 \ge 0
\]
Using the assumption $ a \plus{} b \plus{} c \equal{} 1,ab \plus{} bc \plus{} ca \equal{} q,abc \equal{} r$, the last inequality is equivalent to
\[ 3r(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \ge 0
\]
But it's trivial since
\[ 3r(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \ge q^2(3q \minus{} 10) \plus{} q^2(11 \minus{} 6q) \equal{} q^2(1 \minus{} 3q) \ge 0
\]
We have done. :)[/quote]
You aldrealy use computer to soluted it.!

1. If $ a,b,c \ge 0$ then
\[ (a \plus{} b \plus{} c)^6 \ge 1728abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\]
It very easy, I make now! if false, i hope ....
The my solution:

Assume $ p \equal{} a \plus{} b \plus{} c \equal{} 1$ $ \minus{} > q \equal{} ab \plus{} bc \plus{} ca \leq \frac {1}{3}$

We must prove that:

$ 1 \geq 1728abc(a \minus{} b)(b \minus{} c)(c \minus{} a)$

Only with $ (a \minus{} b)(b \minus{} c)(c \minus{} a) > 0$

We know:

$ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} \sqrt {p^2q^2 \minus{} 4q^3 \plus{} 2(9pq \minus{} 4p^3)r \minus{} 27r^2}$  with $ p \equal{} 1$ and $ abc \equal{} r$

The inequality is equivalent to

$ 1 \geq 1728r \sqrt {q^2 \minus{} 4q^3 \plus{} 2(9q \minus{} 2p^3)r \minus{} 27r^2}$

Put $ k \equal{} 1728$ to it's small!

or $ f(r) \equal{} \minus{} 27k^2r^4 \plus{} 2k^2(9q \minus{} 2)r^3 \plus{} k^2(q^2 \minus{} 4q^3)r^2 \minus{} 1 \leq0$

We only prove with $ q \leq \frac {1}{4}$, if no we easy have it.

$ f'(r) \equal{} \minus{} 108k^2r^3 \plus{} 6k^2(9q \minus{} 2)r^2 \plus{} 2k^2(q^2 \minus{} 4q^3)$

$ f''(r) \equal{} \minus{} 324k^2r^2 \plus{} 12k^2(9q \minus{} 2)r \plus{} 2k^2(q^2 \minus{} 4q^3)$

$ f'''(r) \equal{} \minus{} 648k^2r \plus{} 12k^2(9q \minus{} 2) \leq 0$

$ \minus{} > f''(r) \leq f(0) \equal{} \minus{} 324k^2 \plus{} 12k^2(9q \minus{} 2) \plus{} 2k^2(q^2 \minus{} 4q^3)\leq 0$   (because $ k \equal{} 1728$ and $ q\leq \frac {1}{4}$)

$ \minus{} > f'(r) \leq f(0) \equal{} 0$

$ \minus{} > f(r) \leq f(0) \equal{} \minus{} 1 \leq 0$

We have done!

I don't think you can assume that $ a\plus{}b\plus{}c\equal{}1$ because above inequaliy is [b]cyclic[/b] (of course it means that above one is not [b]symetric[/b]) :wink:

[quote="zaizai-hoang"]I don't think you can assume that $ a \plus{} b \plus{} c \equal{} 1$ because above inequaliy is [b]cyclic[/b] (of course it means that above one is not [b]symetric[/b]) :wink:[/quote]
it ^6, not need cyclic!

[quote="vipCD_A1"]$ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} \sqrt {p^2q^2 \minus{} 4q^3 \plus{} 2(9pq \minus{} 4p^3)r \minus{} 27r^2}$  with $ p \equal{} 1$ and $ abc \equal{} r$
[/quote]
I haven't read your solution yet but I think you have some mistakes here since
\[ (a \minus{} b)^2(b \minus{} c)^2(c \minus{} a)^2 \equal{} p^2q^2 \minus{} 4q^3 \plus{} 2(9pq \minus{} 2p^3)r \minus{} 27r^2
\]

[quote="Ly Tieu Long"][quote="can_hang2007"]1) (VasC) Let $ a,b,c \ge 0,$ prove that
\[ \sum \frac {\sqrt {a^2 \plus{} 2bc}}{b \plus{} c} \ge \frac {3\sqrt {3}}{2}
\]
:)

2) For $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4.$ Prove that
\[ a \plus{} b \plus{} c \le 3 \plus{} \frac {1}{4}\cdot \max ((a \minus{} b)^2,(b \minus{} c)^2,(c \minus{} a)^2)
\]
:)
Equality holds for $ a \equal{} b \equal{} c \equal{} 1$ and $ a \equal{} b \equal{} 2,c \equal{} 0.$ :)[/quote]

Hello my brother, see here:
2) For $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4.$ Prove that
\[ a \plus{} b \plus{} c \le 3 \plus{} \frac {1}{4}\cdot \max ((a \minus{} b)^2,(b \minus{} c)^2,(c \minus{} a)^2)
\]
I think it must be:
2) For $ a,b,c \ge 0,ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4.$ Prove that
\[ a \plus{} b \plus{} c \plus{} \frac {1}{4}\cdot \max ((a \minus{} b)^2,(b \minus{} c)^2,(c \minus{} a)^2) \le 3
\]
[/quote]
No, The orginal problem is true, my younger sister. :) The problem you posted is not true since we always have $ a\plus{}b\plus{}c \ge 3$ by the AM-GM. :)
Best regard.

[quote="can_hang2007"][quote="shfdfzhjj"][quote="can_hang2007"]For $ a,b,c \ge 0$ then
\[ \sum \sqrt [3]{\frac {a^2 \plus{} bc}{b^2 \plus{} c^2}} \ge 2 \plus{} \frac {1}{\sqrt [3]{2}}
\]
:)

[hide="to tanpham90"] I would like to see your solution :), (without using computer for this inequality). :)[/hide][/quote]

Could your give me some hints？
I think all inequalities you have posted are very hard and nice.
I wonder if there exisit nice solution.[/quote]
[hide="hint"] If $ a \ge b \ge c$ then
\[ \frac {a^2 \plus{} bc}{b^2 \plus{} c^2} \plus{} \frac {b^2 \plus{} ca}{c^2 \plus{} a^2} \plus{} \frac {6(a^2 \plus{} bc)}{a^2 \plus{} c^2} \ge \frac {4(a^2 \plus{} b^2)}{c^2 \plus{} ab}
\]
:)
[/hide][/quote]

Dear can_hang2007,

can you please give us another hint, as I am in trouble with this very hard inequality :(

Thank you very much.

[quote="can_hang2007"]For $ a,b,c \ge 0$ then
\[ \sum \sqrt [3]{\frac {a^2 \plus{} bc}{b^2 \plus{} c^2}} \ge 2 \plus{} \frac {1}{\sqrt [3]{2}}
\]
:)
[/quote]
Now, let I post my proof for it. It seems not nice so I am looking for another proof. :)
Here is my proof
Assume that $ a \ge b \ge c$, then we will show that
\[ \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}} \ge \max\left\{ 2, \sqrt[3]{\frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}}\right\}\]
Indeed, we have
\[ \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}} \ge 2\sqrt[6]{\frac{(a^2\plus{}bc)(b^2\plus{}ca)}{(b^2\plus{}c^2)(a^2\plus{}c^2)}} \ge 2\sqrt[6]{\frac{(a^2\plus{}c^2)(b^2\plus{}c^2)}{(b^2\plus{}c^2)(a^2\plus{}c^2)}}\equal{}2\]
So, to prove our claim, we need to show that
\[ \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}} \ge \sqrt[3]{\frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}}\]
\[ \Leftrightarrow \frac{a^2\plus{}bc}{b^2\plus{}c^2} \plus{}\frac{b^2\plus{}ca}{c^2\plus{}a^2} \plus{}3\sqrt[3]{\frac{(a^2\plus{}bc)(b^2\plus{}ca)}{(a^2\plus{}c^2)(b^2\plus{}c^2)}} \left( \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}}\right) \ge \frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}\]
We have
\[ \frac{b^2\plus{}ca}{b^2\plus{}c^2} \minus{}\frac{a^2\plus{}bc}{a^2\plus{}c^2} \equal{}\frac{c(a\minus{}b)(a^2\plus{}b^2\plus{}c^2\plus{}ab\minus{}ac\minus{}bc)}{(a^2\plus{}c^2)(b^2\plus{}c^2)} \ge 0\]
Thus, by AM-GM Inequality, we have
\[ 3\sqrt[3]{\frac{(a^2\plus{}bc)(b^2\plus{}ca)}{(a^2\plus{}c^2)(b^2\plus{}c^2)}} \left( \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}} \plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}}\right) \ge  6\sqrt{\frac{(a^2\plus{}bc)(b^2\plus{}ca)}{(a^2\plus{}c^2)(b^2\plus{}c^2)}} \ge \frac{6(a^2\plus{}bc)}{a^2\plus{}c^2}\]
It suffices for us to show that
\[ \frac{a^2\plus{}bc}{b^2\plus{}c^2}\plus{}\frac{b^2\plus{}ca}{c^2\plus{}a^2}\plus{}\frac{6(a^2\plus{}bc)}{a^2\plus{}c^2} \ge \frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}\]
which is equivalent to
\[ f(c)\plus{}g(c) \ge  0\]
where
\[ f(c) \equal{} (a \plus{} 7b)c^5 \plus{} 3(a^2 \minus{}b^2)c^4 \plus{} 2(a \plus{} b)(a \plus{} 3b)bc^3 \ge 0\]
\[ g(c) \equal{} (a \minus{} b)(3b^3 \plus{} 2ab^2 \plus{} 4a^2b \minus{} 3a^3)c^2 \plus{} (b^2a^3 \plus{} 6b^4a \plus{} a^2b^3)c \plus{} ab(a \minus{} b)^4\]
If $ 3b^3 \plus{} 2ab^2 \plus{} 4a^2b \minus{} 3a^3 \ge 0$, then clearly $ g(c) \ge 0$.
If $ 3b^3 \plus{} 2ab^2 \plus{} 4a^2b \minus{} 3a^3 \le 0$, then $ g(c)$ is concave for all $ c \in [0,b]$, thus
\[ g(c) \ge \min\{g(0),g(b)\}\]
But
\[ g(0)\equal{} ab(a \minus{} b)^4 \ge 0\]
\[ g(b) \equal{} b \left[ \frac{1}{4}(a\minus{}b)[(2a^2\minus{}6ab\minus{}b^2)^2 \plus{}43b^4]\plus{}8b^5\right] \ge 0\]
THus, our claim is proved.
Now, back to our problem
We will consider 2 cases
Case 1. If $ \frac{a^2\plus{}b^2}{c^2\plus{}ab} \le 2$, then we have
\[ \sqrt[3]{\frac{c^2\plus{}ab}{a^2\plus{}b^2}} \ge \frac{1}{\sqrt[3]{2}}\]
and
\[ \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}}\plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}} \ge 2\]
\[ \Rightarrow \;\; \text{done}\]
Case 2. If $ \frac{a^2\plus{}b^2}{c^2\plus{}ab} \ge 2$, then we have
\[ \sqrt[3]{\frac{a^2\plus{}bc}{b^2\plus{}c^2}}\plus{}\sqrt[3]{\frac{b^2\plus{}ca}{c^2\plus{}a^2}} \ge \sqrt[3]{\frac{4(a^2\plus{}b^2)}{c^2\plus{}ab}}\equal{}4\sqrt[3]{4}x\]
\[ \Leftrightarrow LHS \ge \sqrt[3]{4} x\plus{} \frac{1}{x} \ge 2\plus{}\frac{1}{\sqrt[3]{2}}\]
(since $ x \ge \sqrt[3]{2}$)
We have done.
:)
Equality holds if and only if $ (a,b,c)\sim (1,1,0).$

thank you very much,but you haven't post the solution to it :( ,it is very difficult

$\sum \frac{a}{b}\ge 3+\frac{6(a-c)^2}{(a+b+c)^2}$

[hide]another similar inequality $3(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+2\ge \frac{33(a^2+b^2+c^2)}{(a+b+c)^2}$

it is stronger than $2(\frac{a}{b}+\frac{b}{c}+\frac{c}{a})+1\ge \frac{21(a^2+b^2+c^2)}{(a+b+c)^2}$

but it is also hard[/hide]

[quote="can_hang2007"]For any $ a,b,c>0$
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}
\]
:)[/quote]
My ugly proof:
1. $a=\min\{a,b,c\}$, $b=a+x$ and $c=a+y$.
Since $\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}- 3=\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1=\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}$, we obtain:
$\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}\Leftrightarrow\frac{x^2}{a(a+x)}+\frac{y^2-xy}{a(a+y)}\geq\frac{6y^2}{(3a+x+y)^2}\Leftrightarrow$
$\Leftrightarrow3(3x^2-3xy+y^2)a^3+3x(3x^2+y^2)a^2+$
$+(x^4+x^3y+6x^2y^2+xy^3+y^4)a+xy^2(x+y)^2\geq0$, which is true.
2. $b=\min\{a,b,c\}$, $a=b+x$ and $c=b+y$.
Hence, $\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}\Leftrightarrow3(x^2+xy+y^2)b^3+3xy(5x+2y)b^2+$
$+(x^4+x^3y+18x^2y^2-5xy^3+y^4)b+x^2y(x+y)^2\geq0$, which is true.
3. $c=\min\{a,b,c\}$, $a=c+x$, $b=c+y$ and $x=ty$.
Hence, $\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}\Leftrightarrow3(x^2-3xy+3y^2)c^3-3y(2x^2-3xy-2y^2)c^2+$
$+(x^4-5x^3y+6x^2y^2+7xy^3+y^4)c+xy^2(x+y)^2\geq0$.
Since $x^4-5x^3y+6x^2y^2+7xy^3+y^4\geq0$, it remains to prove our inequality for $2x^2-3xy-2y^2\geq0$, which gives $t\geq2$.
Id est, enough to prove that
$\left(3y(2x^2-3xy-2y^2)\right)^2-12(x^2-3xy+3y^2)(x^4-5x^3y+6x^2y^2+7xy^3+y^4)\leq0$, which is
$t(4t^5-32t^4+84t^3-68t^2-11t+36)\geq0$, which is true for $t\geq2$.
What is your proof, can_hang2007? Thank you!

[quote="can_hang2007"]
2. If $ a,b,c \ge 1/2,abc \equal{} 1$ then
\[ \frac {1}{a \plus{} 2b} \plus{} \frac {1}{b \plus{} 2c} \plus{} \frac {1}{c \plus{} 2a} \le 1
\]
[/quote]
Let $a=\frac{4x+2y+z}{4y+2z+x}$, $b=\frac{4y+2z+x}{4z+2x+y}$ and $c=\frac{4z+2x+y}{4x+2y+z}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that:
$\sum_{cyc}(64x^6+248x^5y+1720x^5z+1360x^4y^2+6402x^4z^2-2098x^4yz+$
$+7062x^3y^3-3743x^3y^2z-6320x^3z^2y-4695x^2y^2z^2)\geq0$, which is obviously true.
What is your proof, Can?

[quote="can_hang2007"][hide]1. If $ a,b,c \ge 0$ then
\[ (a \plus{} b \plus{} c)^6 \ge 1728abc(a \minus{} b)(b \minus{} c)(c \minus{} a)
\][/hide]2. If $ a,b,c \ge 1/2,abc \equal{} 1$ then
\[ \frac {1}{a \plus{} 2b} \plus{} \frac {1}{b \plus{} 2c} \plus{} \frac {1}{c \plus{} 2a} \le 1
\]3. If $ a,b,c \ge 0$ then
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \ge \frac {3}{2} \plus{} \frac {27}{16}\cdot \frac {(a \minus{} c)^2}{(a \plus{} b \plus{} c)^2}
\][/quote]
contrast .
$ \frac{a}{2c\plus{}b} \plus{} \frac{b}{2a\plus{}c} \plus{} \frac{a}{2b\plus{}c} \geq 1(a,b,c>0)$

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=1362745
Given that $ a,b,c \geq 0$, Prove that:
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\geq \frac {3}{2} \plus{} 2\sum_{cyc}\left(\frac {a \minus{} b}{a \plus{} b \plus{} 2c}\right)^2\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=1365454

[quote=can_hang2007]1) (VasC) Let $ a,b,c \ge 0,$ prove that
\[ \sum \frac{\sqrt{a^2\plus{}2bc}}{b\plus{}c} \ge \frac{3\sqrt{3}}{2}
\]
:)

2) For $ a,b,c \ge 0,ab\plus{}bc\plus{}ca\plus{}abc\equal{}4.$ Prove that
\[ a\plus{}b\plus{}c \le 3 \plus{}\frac{1}{4}\cdot \max ((a\minus{}b)^2,(b\minus{}c)^2,(c\minus{}a)^2)
\]
:)
Equality holds for $ a\equal{}b\equal{}c\equal{}1$ and $ a\equal{}b\equal{}2,c\equal{}0.$ :)[/quote]

[img]http://s7.sinaimg.cn/middle/006ptkjAzy75Sly0p02e6&690[/img]

[quote=can_hang2007]Problem. Prove the inequality for $ x,y,z \in \mathbb{R}$
\[ \sum \frac{x^2\minus{}yz}{4x^2\plus{}4y^2\plus{}z^2} \ge 0
\]
My solution.
By expanding, the inequality is equivalent to
\[ 4\sum x^6\minus{} 4\sum x^5y \minus{}16\sum xy^5 \plus{} 33\sum x^4y^2 \plus{}24\sum x^2y^4 \minus{}20 \sum x^3y^3 \minus{}4\sum x^4yz \minus{}20\sum x^3y^2z \minus{}17\sum x^3yz^2 \plus{}60x^2y^2z^2 \ge 0
\]
Now, setting $ x\equal{}4a\plus{}2b\plus{}c,y\equal{}4b\plus{}2c\plus{}a,z\equal{}4c\plus{}2a\plus{}b$, the inequality becomes
\[ 22\sum a^4(b^2\plus{}c^2) \plus{}32\sum a^3b^3 \minus{}16\sum a^4bc \minus{}22 \sum a^2b^2c(a\plus{}b) \minus{}48a^2b^2c^2 \ge \minus{}3(\sum a^4b^2 \minus{}\sum a^3bc^2)
\]
By the AM-GM inequality, we have  $ \minus{}3(\sum a^4b^2 \minus{}\sum a^3bc^2) \le 0$, hence it suffices to prove
\[ 22\sum a^4(b^2\plus{}c^2) \plus{}32\sum a^3b^3 \minus{}16\sum a^4bc \minus{}22 \sum a^2b^2c(a\plus{}b) \minus{}48a^2b^2c^2 \ge 0
\]
or
\[ 11\sum a^4(b^2\plus{}c^2) \plus{}16\sum a^3b^3 \minus{}8\sum a^4bc \minus{}11 \sum a^2b^2c(a\plus{}b) \minus{}24a^2b^2c^2 \ge 0
\]
Using the assumption $ a\plus{}b\plus{}c\equal{}1,ab\plus{}bc\plus{}ca\equal{}q,abc\equal{}r$, the last inequality is equivalent to
\[ 3r(3q\minus{}10) \plus{}q^2(11\minus{}6q) \ge 0
\]
But it's trivial since
\[ 3r(3q\minus{}10)\plus{}q^2(11\minus{}6q) \ge q^2(3q\minus{}10)\plus{}q^2(11\minus{}6q)\equal{}q^2(1\minus{}3q) \ge 0
\]
We have done. :)[/quote]
Cauchy inequality

[quote=can_hang2007]Problem. Prove the inequality for $ x,y,z \in \mathbb{R}$
\[ \sum \frac{x^2\minus{}yz}{4x^2\plus{}4y^2\plus{}z^2} \ge 0\][/quote]

[img]http://s8.sinaimg.cn/middle/006ptkjAzy75YCoz6SP87&690[/img]

[quote=luofangxiang][img]http://s8.sinaimg.cn/middle/006ptkjAzy75YCoz6SP87&690[/img][/quote]
It's similar Vo Quoc Ba Can solution in his book. Moreover, we have
\[\sum \frac{x^2 - yz}{2kx^2+k^2y^2+z^2} \geqslant 0,\]
for all $k > 0.$

[quote=can_hang2007]4. For any $ a,b,c,d \ge 0$
\[ \sum_{sym} \frac{1}{a^2\plus{}b^2} \ge \frac{81}{2(a\plus{}b\plus{}c\plus{}d)^2}
\]
:)[/quote]

[img]http://s15.sinaimg.cn/middle/006ptkjAzy75YG170Fo1e&690[/img]

[quote=can_hang2007]For any $ a,b,c>0$
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge 3\plus{}\frac{6(a\minus{}c)^2}{(a\plus{}b\plus{}c)^2}
\]
:)[/quote]

[hide=Sol]We will prove the resault:
$\textbf{Lemma}\\ \text{For any } a,b,c>0, \text{we have: } \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3+\frac{4[(b-c)^{2}+(c-a)^{2}+(a-b)^{2}]}{(a+b+c)^{2}}$
WLOG, assume that $b$ is between $a$ and $c$.
$\Rightarrow LHS\geq 1+\frac{c}{a}+\frac{a}{c}=3+\frac{(c-a)^{2}}{ca}$
But $ca\leq \frac{1}{8}.min\lbrace (2c+a)^{2}, (2a+c)^{2} \rbrace\leq \frac{(a+b+c)^{2}}{8}$
$\Rightarrow LHS\geq 3+\frac{8(a-c)^{2}}{(a+b+c)^{2}}=RHS+\frac{8(a-b)(b-c)}{(a+b+c)^{2}}\geq RHS$
Now, by Cauchy-Schwarz inequality:
$3+\frac{4[(b-c)^{2}+(c-a)^{2}+(a-b)^{2}]}{(a+b+c)^{2}}\geq 3+\frac{2(a-b+b-c)^{2}+4(c-a)^{2}}{(a+b+c)^{2}}=3+\frac{6(a-c)^{2}}{(a+b+c)^{2}}$ ($Q.E.D$)
Equality holds at $a=b=c$. $\square$

[quote=Nguyenhuyen_AG][quote=VMF-er]
$\text{For any } a,b,c>0, \text{we have: } \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3+\frac{4[(b-c)^{2}+(c-a)^{2}+(a-b)^{2}]}{(a+b+c)^{2}}$
[/quote]
Are you sure, check $a = \frac 35,$ $b = \frac 12,$ $c = 1$ [img]https://voz.vn/styles/next/xenforo/smilies/popo/eek.png?v=01[/img][/quote]

Oops :lol: My proof is wrong too  :blush:

With same condition$,$ the following inequality is true!
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\geqslant 3+\dfrac{3\sqrt[3]{2} \sum (a-b)^2}{(a+b+c)^2}$$
[b]Note.[/b] $3\sqrt[3]{2}$ is also the best constant here!


AMC
AIME
trigonometry
geometry
ARML
AMC 10

[quote="jb05"]144... I had a careless mistake on the brute force #25 (AMC10). Yes, I know you can reduce it to 12 cases, and in fact I did that during the test. I just incorrectly added up one of the S(n)'s. I hate problems like that.

Any perfects on the 10a?[/quote]

Same here  :(

I had for some reason convinced myself that there weren't any in the 1980s other than 1980...

I know at my school, no one else qualified for the AIME through the AMC 10A (only one other person got over 100, I think).

two guys qualified through the 10 at my school.

Not me.

If someone here has a copy of the booklet, could they scan it and PM it to me? Thanks.

[quote="b-flat"][quote="jb05"]144... I had a careless mistake on the brute force #25 (AMC10). Yes, I know you can reduce it to 12 cases, and in fact I did that during the test. I just incorrectly added up one of the S(n)'s. I hate problems like that.

Any perfects on the 10a?[/quote]

Same here  :(

I had for some reason convinced myself that there weren't any in the 1980s other than 1980...
[/quote]

I did the same...

AND missed 23 due to misbubbling!! :(

How can you possibly know you misbubbled something a day after the test?  :huh:

My teacher kept photocopies of the answer sheets... I remember getting 4, then bubbled C... (which was 6)

[quote="jb05"]Any perfects on the 10a?[/quote]

There was one last year. Possible candidates for this year:

  John Berman  ()
  Tianyi Jiang ()
  Bryce  Taylor ()
  Brendan Fletcher ?
  Jenny Chen ?
  Daniel Vitek ?
  ??????

I had already determined those actually... Vitek isn't taking it. Just the 12a, and he got a 123.

I love all the stuff about... so and so (failed). It sounds very dramatic.

Apparently, Jenny Chen got a perfect.

[quote="jb05"]I had already determined those actually... Vitek isn't taking it. Just the 10a, and he got a 120-something.

I love all the stuff about... so and so (failed). It sounds very dramatic.

Apparently, Jenny Chen got a perfect.[/quote]

Nice.  Where did you here that?  I know she is quite good...

I guess that means jb05 & I won't win then :(

Also, who is Tianyi Jiang?  I don't remember  :blush:

I will see if I can ask Brendan Fletcher (I talk to him at least once a week) what he got.

I got 22 and missed 3, for a total of 132.
2 of them I missed because of careless errors... :(

[quote="b-flat"]
Also, who is Tianyi Jiang?  I don't remember  :blush:
[/quote]

omg

you can guess who it is now

Evidently, the substantial efforts to provide socializing opportunities for the NC math elite have failed.

Tianyi Jiang has placed k posts on this thread prior to this post.  Solve the following to find k.

Find the smallest integer value of k greater than two such that the product of the factors of 2008 can be written in the form n^k where n is a positive integer.

[quote="jb05"]Can somebody please explain... Who's Tianyi Jiang???[/quote]

I don't even live in North Carolina, let alone know anyone there or have ever stepped my foot on the planet.

I have implied from previous posts that not_trig is Tianyi Jiang...

I wonder why something is called common sense if people don't have it  :rotfl:

Ah. My apologies. You see, I thought not_trig was Damien Jiang, so while the hints were rather clear, I was sure he wasn't Tianyi. Apparently, Damien=Tianyi, which makes things a lot clearer. But I actually do know who he is, I just can't keep track of two names at once...

oh wow.

That's pretty nice... two names to be known as....

 :lol:

Damien is my middle name, Tianyi my first... :rotfl:

2007 AMC B Registration information for NC:

Total schools  : 28
AMC 10 Bundles:   69
AMC 12 Bundles:    73

Nobody wants to talk about the AMC 10B/12B?

No! Obviously not, especially me, since I failed.

[quote="not_trig"]I failed.[/quote]

Failed again? You failed twice within one month ?

I do not think that 138 for AMC 10 and 125 for AMC 12 are "failed" scores.

OK, for the 12B, I made 4 stupid errors & got a 120 (I think).  Would have gotten 141 without.

125? I wish. I definitely completely FAILED.

not_trig:  Sometimes it's difficult to judge how well you've done until the score distributions are released.  The AMC's were tough this year - average scores will probably be down 10-15 points from last year.  Many students are reporting lower scores this year than last year.  There may be far fewer high scores (125+) this year in NC than in the recent past.

The Winston-Salem area math teachers seem to think there will be very few AIME qualifiers locally.  Of course, Winston-Salem isn't noted for producing top math talent.  We'll probably have just one AIME qualifier at Hanes this year (vs. 4 last year.)

Since you qualified for the AIME, the AMC score doesn't really matter.  We'll probably have fewer than 200 AIME participants in NC this year.  I'll bet very few of these are freshmen or below.

Find a good place to qualify for the state math contest.  It'll be fun.

Yeah, I need to focus on AIME and State Math Contest instead of how I failed AMCs and ARML practice... :D

Mr. Jeremy Hahn--Why didn't you guess on the last question???

North Carolina!!

Welcome to the North Carolina forum!!

Feel free to ask any questions about anything  :)

Also, good job on the 12B.

jb05 - please edit.  i got a 123 in amc12A.  a 120something on 10A would be almost embarrasing, and a more than 20point drop.  thx

Oh, heh. I actually knew that, just a typo. I guess I should let people post for themselves what they got...

It's too late to edit.

nah - i don't care if you know what i got, or even if you post it for me.  at least it's fixed.  all I really care about now is doing well on USAMO.

Hm, I know 4 people from my school [Ligon MS] qualified for AIME through 10A: Arjun Bhattacharya, Calvin Deng, James Ding, Nick Tobey.

I do apologize profusely for the profanity I posted on February 16 on this thread. Because of this, I may not post for the next 3 months. B-flat, please edit out the profanity in the post. Thank you.

[quote="JennyC"]I got 129 in AMC 12B.[/quote]

why does she have 0 posts?

Edit: Sorry forgot about the post count change...

[quote="MysticTerminator"]dang an old thread that HAS NOT YET BEEN LOCKED THIS IS A CRAZY PLACE TO BE, GUYS[/quote]

GOOD POINT!


I have Volume 2. In what order should I study it (so as to go from lowest level to highest level; it seems most say that the chapters are arranged well by topic but not difficultly)?

That pretty much depends on what you already know in each subject.  Glance at each of the chapters and start with the one you understand the most words in?


The freshmen like me on the physics team in my high school are being taught calculus by upperclassmen in the club. Where would be a good place here to ask about basic calculus?

Try the Calculus forum: http://www.artofproblemsolving.com/Forum/index.php?f=296

[url=http://www.artofproblemsolving.com/Forum/index.php?f=150]If it's just simple limits, the Intermediate Forum is fine.
[/url]


i know its a silly one but seriously...

what is $ min{a^3\plus{}b^3\plus{}c^3} \equal{}$???

[quote="sfsreporter"]i know its a silly one but seriously...

what is $ min{a^3 \plus{} b^3 \plus{} c^3} \equal{}$???[/quote]
Not enough to tell you an answer
What is the condition of $ a,b,c$.If not The result is $ \minus{}\infty$

hello, the searched minimum of $ a^3\plus{}b^3\plus{}c^3$  under the condition $ a\geq0,b\geq0,c\geq0$ is zero.
Sonnhard.

No. Minimum Value by definition is independent of the variables. However if you ask for $ \min \left(\frac {a^3 \plus{} b^3 \plus{} c^3}{abc}\right)$ for positive reals $ a,b,c$ the answer is $ 3$.

sfsreporter you have a wrong understanding of the inequalities.
AM greater than or equal to GM doesnt necessarily give you the minimum.


trigonometry
geometry
circumcircle
inequalities
inequalities unsolved

Let triangle $ ABC$ .Prove that
             $ \frac{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}{1+\sin\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}}< \sqrt2$

[quote="Nbach"][color=darkred][b]Let triangle $ ABC$. Prove that[/b][/color] $ \frac{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}{1+\sin\frac{A}{2}.\sin\frac{B}{2}.\sin\frac{C}{2}}<\sqrt 2\ .$[/quote]
[color=darkblue][b]Proof.[/b] Denote the incircle $ C(I,r)$ and the $ A$- exincircle $ C(I_{a},r_{a})$ of $ \triangle ABC$. Observe that

$ \prod\sin \frac{A}{2}=\frac{(p-a)(p-b)(p-c)}{abc}=\frac{pr^{2}}{4Rpr}=\frac{r}{4R}$ $ \implies$ $ 1+\prod\sin\frac{A}{2}=\frac{4R+r}{4R}$.
Let $ [AA']$ be a diameter in the circumcircle of the triangle $ ABC$. Then $ A'I=A'I_{a}=A'B=A'C=2R\sin\frac{A}{2}$ a.s.o. and $ \sum\sin\frac{A}{2}=\frac{1}{2R}\cdot\sum IA'$.
The our inequality becomes : $ \frac{\frac{1}{2R}\cdot \sum IA'}{\frac{4R+r}{4R}}<\sqrt 2$ $ \Longleftrightarrow$ $ \sqrt 2\cdot\sum IA'\le 4R+r$ $ \Longleftrightarrow$  $ \boxed{\sum II_{a}<\sqrt 2\cdot (4R+r)}$, because $ II_{a}=2\cdot IA'$.[/color]

Here [b]Campos[/b] remarked correctly a mistake.  I deleted it and I hope come back soon here !

[quote="campos"]are you sure about that? the equality holds for $ (A,B,C)=(90,90,0)$ so the bound $ \frac{4}{3}$ isn't correct.[/quote]Yes, you are right ! I made a mistake here : .... But $ II_{a}\le r+r_{a}$ ....
Correctly is ... But $ II_{a}\ge r+r_{a}$ ... Thank you for your understanding and sorry for it.

For the sidelenghts of the triangle we use the Ravi substitutions $ a=y+z$,$ b=z+x$,$ c=x+y$ with $ x,y,z\geq 0$.Now we have $ \sin{\frac{A}{2}}=\sqrt{\frac{yz}{(x+y)(x+z)}}$ and the similar.The inequlity we want to prove is $ \sum_{cyc}\sqrt{\frac{yz}{(x+y)(x+z)}}\leq \frac{\sqrt{2}(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)}$.From CBS we have that
$ \left(\sum_{cyc}\sqrt{\frac{yz}{(x+y)(x+z)}\right)^{2}\leq \left(\sum_{cyc}xy\right)\left(\sum_{cyc}\frac{1}{(x+y)(x+z)}\right)}$.After we intercalate this in the given inequality we just have to prove $ (x+y)(y+z)(z+x)\leq (x+y+z)(xy+yz+zx)$ which is trivial. :)

$ Solution:$
We will use the following known identities:

For the acute angled triangle abc
(1.) $ \sum{sin^{2}a}=2+2\prod{cosa}$ $ \Rightarrow $ $ \sum{sin^{2}a}>2$ (2.).
(3.) $ \sum{sin^{2}a}\geq (\sum{cosa})^{2}$ (Problem 10.8-c Н. М. Седракян: Геометрические неравентва, Ереван 2004).

Let $ cyc(a=90-A/2)$ $ \Rightarrow $ our inequality is equivalent to:

$ \frac{\sum{cosa}}{1+\prod{cosa}}<\sqrt{2}$, using (1) it's enough to prove $ \sum{\sqrt{2}cosa}<\sum{sin^{2}a}$
on the other hand, by (2.) and (3.) we have $ (\sum{sin^{2}a})^{2}>2\sum{sin^{2}a}\geq 2(\sum{cosa})^{2}$
$ \Rightarrow $ $ \sum{sin^{2}a}>\sum{\sqrt{2}cosa}$, which completes our proof.

[quote="Nbach"][color=darkred][b]Let triangle $ ABC$. Prove that[/b][/color] $ \boxed{\frac{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}{1+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}<\sqrt 2}\ \ (*)$.[/quote]
[color=darkblue][b]Proof.[/b] Denote the incircle $ C(I,r)$ and the $ A$- exincircle $ C(I_{a},r_{a})$ of $ \triangle ABC$. Observe that

$ \prod\sin\frac{A}{2}=\frac{(p-a)(p-b)(p-c)}{abc}=\frac{r}{4R}$ $ \implies$ $ 1+\prod\sin\frac{A}{2}=\frac{4R+r}{4R}$ and the our inequality $ (*)$ becomes $ \boxed{\sum\sin\frac{A}{2}<\frac{(4R+r)\sqrt 2}{4R}}\ \ (**)$.

$ \sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}=\sqrt{\frac{a(p-a)(p-b)(p-c)}{abc(p-a)}}=$ $ \sqrt{\frac{apr^{2}}{4Rpr\cdot (p-a)}}=\sqrt{\frac{ar}{4R(p-a)}}$ $ \implies$ $ \boxed{\sum\sin\frac{A}{2}=\sqrt{\frac{r}{4R}}\cdot\sum\sqrt{\frac{a}{p-a}}}\ \ (2)$.

Apply [b]C.B.S.'s inequality[/b] : $ \left(\sum\sqrt{\frac{a}{p-a}}\right)^{2}\le\sum a\cdot\sum\frac{1}{p-a}=$ $ 2p\cdot\frac{r(4R+r)}{pr^{2}}$ $ \implies$ $ \boxed{\sum\sqrt{\frac{a}{p-a}}\le\sqrt{\frac{2(4R+r)}{r}}}\ \ (3)$.

Thus, from the identity $ (2)$ and the inequality $ (3)$ obtain $ \sum\sin\frac{A}{2}\le\sqrt{\frac{r}{4R}}\cdot\sqrt{\frac{2(4R+r)}{r}}$ $ \implies$ $ \boxed{\sum\sin\frac{A}{2}\le\sqrt{\frac{4R+r}{2R}}}\ \ (4)$.

Now see the required inequalitiy $ (**)$ and the inequality $ (4)$. Is sufficiently to prove ${ \sqrt{\frac{4R+r}{2R}}<\frac{(4R+r)\sqrt 2}{4R}}$.

Indeed, this inequality is equivalently with $ 4R < 4R+r$, i.e. $ r > 0$, what is truly.

In conclusion,  $ \boxed{\ \frac{\sin\frac{A}{2}+\sin\frac{B}{2}+\sin\frac{C}{2}}{1+\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}\le\sqrt{2-\frac{2r}{4R+r}}<\sqrt 2\ }$.[/color]


You are given a sequence of real numbers $(a_n)_{n \geq 1}$. Let $L=\{t \in \overline{\mathbb{R}} ~ | ~ \textrm{there is a subsequence convergent to} ~ t ~ \}$. Prove that \[ \lim_{n \to \infty} \left( \sup_{k \geq n} a_k \right) = \sup L .  \]

The definition (on least on mathworld) is: $\lim \sup a_n = \lim_{n \to \infty} \left( \sup_{k \geq n} a_k \right)$. in class, I was thought the other one. ($\lim \sup a_n = \sup L$) i was asking for a proof of their equivalence

Suppose $b_n = \sup_{k\geq n} a_k$ and $M = \lim_{n \to \infty} b_n$. We assume that $M$ is finite. ($\{b_n\}$ is decreasing, in fact). We'll show that $\sup L = M$.

First we claim that $M\in L$. From the definition of $b_n$, we can choose $N_1<N_2<\cdots$ such that $|b_k - a_{N_k}| < \frac 1k$. Thus $|a_{N_k} - M| < |b_k - a_{N_k}|+|b_k-M|<\frac 1k+|b_k-M|$. For a given $\epsilon>0$, we can find $K$ large enough such that it holds both $|b_k-L|<\epsilon/2$ and $\frac 1k<\epsilon/2$ for all $k>K$. Then $|a_{N_k} - M|<\epsilon$ for $k>K$. Thus $a_{N_k}$ converges to $M$.

Now we claim that $M$ is an upperbound of $L$. If $M'\in L$ and $M'>M$, then there exists $a_{m_k}\to M'>M$. Then $a_{m_k}>\alpha>M$ for $k$ large enough. Thus $b_n>\alpha$, $\lim b_n\geq \alpha>M$. A contradiction.

Therefore, $\sup L=M$.

[quote="liyi"]From the definition of $b_n$, we can choose $N_1<N_2<\cdots$ such that $|b_k - a_{N_k}| < \frac 1k$.[/quote]

can we always do that? what if we had $1,1,1,1,1,3,1,1,1,1,...$ (after that an inf string of ones)?

could this become a problem    :?   :blush:

[quote="Kent Merryfield"][quote="perfect_radio"]what if we had 1,1,1,1,1,3,1,1,1,1,... (after that an inf string of ones)? [/quote]
Then for sufficiently large $k,\,b_k=1.$ The only limit of a subsequence for this sequence would be 1.

When liyi said "( $\{b_n\}$ is decreasing, in fact)" the precise sense of what he meant when he said "decreasing" was "non-increasing."[/quote]

i figured out that by myself, Kent (that it was nonincreasing)

i talking about the procedure he used. how exactly do we know that [quote]We can choose $N_1<N_2<\cdots$ such that $|b_k - a_{N_k}| < \frac 1k$.[/quote] ?

this may be very obvious to you and liyi, but it's not to me :(

liyi is starting with $k.$ Note that both $b_k$ and $N_k$ depend on the same $k.$

But we could do the whole thing more cleanly. Let $L=\lim_{k\to\infty}\left(\sup_{n\ge k}a_n \right)$. Then we can say that, for all $\epsilon>0$, $a_n$ is [b]eventually[/b] less than $L+\epsilon,$ and $a_n$ is greater than $L-\epsilon$ [b]infinitely often.[/b]

More formally, $\forall\,\epsilon>0,\,\exists\,N$ such that if $n>N,$ then $a_n<L+\epsilon$ and $\forall\,N,\,\exists\,n>N$ such that $a_n>L-\epsilon.$

The first is true because $b_k$ is eventually less than $L+\epsilon$ and for $n\ge k,\,a_n\le b_k.$

The second is true because if it were not true, for all sufficiently large $k,$ we would have $b_k\le L-\epsilon,$ since $L-\epsilon$ would be an upper bound for the tail end of the sequence.

Now choose your favorite sequence that tends to zero - say, $\frac1n.$ From the above, we can find $a_{n_1}$ such that $L-1<a_{n_1}<L+1.$ After that, we can find $a_{n_j}$ such that $n_j>n_{j-1}$ and $L-\frac1j<a_{n_j}<L+\frac1j.$

This is pretty much the argument liyi used, but I rearranged the details and the language to my own tastes.

thanks

[quote="Kent Merryfield"]More formally, $\forall\,\epsilon>0,\,\exists\,N$ such that if $n>N,$ then $a_n<L+\epsilon$ and $\forall\,N,\,\exists\,n>N$ such that $a_n>L-\epsilon.$[/quote]

this line really clarified everything


Find all triples of integrers $(a;b;c)$ such that $a+b+c=24$, $a^2+b^2+c^2=210$, $abc=440$.


$ \frac{3\plus{}4}{3\plus{}5}$ does not equal $ \frac{4}{5}$ but I vaguely recall a rule about fractions about adding something to the numerator and adding probably something else to the denominator and getting an equivalent fraction... I imagine that it probably had to do with ratios.

I know that I'm not being very clear, but does anyone have any ideas about what trick I'm thinking about?

if $ \frac{a}{x}\equal{}\frac{b}{y}$, then $ \frac{a\plus{}x}{x}\equal{}\frac{b\plus{}y}{y}$.

$ \frac{a}{x}\equal{}\frac{b}{y}$ so $ \frac{a\plus{}b}{x\plus{}y}$

3/3 + 4/5 is not equal to 4/5...

1st, if their denominator is not similar, get their least common denominator..

in this case, their LCD is 15..5 x 3

15 divided by the denominator x the numerator of the 1st given + do the same thing to the next fraction..

15+12/15..=27/15 or 1 12/15

another solution is..change 3/3 to 5/5..remember, 3/3=1 so to 5/5...i choose 5/5 because it has the same denominator as the 2nd fraction...

therefore,

5/5 + 4/5 = 5+4/5..just add the numerator because their denominator is similar..therefore, the result is 9/5...

as you noticed, my answers are not the same...however, they have the same value..

have you meant that ?

$ \frac {x}{y} \equal{} \frac {k*x}{k*y}$

or may be

$ \frac {x}{y} \equal{} \frac {z}{k} \rightarrow xk \equal{} yz$

The question has been answered by jli. Think about it.

hehe :lol:

if $ \frac {a}{x} \equal{} \frac {b}{y}$ then

$ b \equal{} ka$
$ y \equal{} kx$

so we have $ \frac {a \plus{} b}{x \plus{} y} \equal{} \frac {a \plus{} ka}{x \plus{} kx} \equal{} \frac {a(1 \plus{} k)}{x(1 \plus{} k)} \equal{} \frac {a}{x}$

realy nice  :)

(however if $ x\equal{}\minus{}y$ then you'll have a problem)


A line segment begins at $ (1, 3)$. It is 5 units long and ends at the
point $ (4, x)$ and $ x > 0$. What is the value of $ x$?

The difference of x coordinates is 3, and since the distance is 5, the difference of y coordinates is 4 (3-4-5 triangle). Since the y intercept is positive, the value of x is 3+4=7.


Let $A(n)$ denote the set of all $2n$digit numbers whose decimal representation conisists of exactly $n$ $1$'s and $n$ $0$'s. (no leading $0$ is permitted.) Show that there must be at least one number in the set $A(n)$ that is evenly divisible by $n$.

Say, for $n = 3$, the set $A(3)$ has $5!/(3!*2!) = 10$ numbers.
$A(3) ={100011, 100110, 101100, 111000, 110100, 110010, 110001, 101010, 101001, 100101}$ of which at least one should be divisible by $3$. Here all of them are!!! That's a different story.

Prove it for any postitive number $n$ irrespective of how big or small $n$ is.

I'm having trouble finding a number than works for $n=7$


In how many triangles does the semiperimeter of the triangle equal the triangle's area?

Or is a $ 3\minus{}4\minus{}5$ triangle the only case of this?

[hide]
There are infinitely many triangles. Its all the triangles that have inradius 1. [/hide]

A 3-4-5 triangle is the only [i]right[/i] triangle that works though, as can be seen by letting the sides be $ 2rs, r^2\minus{}s^2, r^2\plus{}s^2$.

A 3-4-5 triangle is the only right triangle with [i]integer sides[/i] that works.

A nice relation between inradii and right triangles is that if the sides are written $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$ where $ c$ is the hypotenuse, then $ y \equal{} r$ is the inradius.  Setting $ y \equal{} 1$ gives

$ (x \plus{} 1)^2 \plus{} (z \plus{} 1)^2 \equal{} (x \plus{} z)^2 \Leftrightarrow$
$ (x \minus{} 1)(z \minus{} 1) \equal{} 2$

which only has the integer solution $ \{ x, z \} \equal{} \{ 2, 3 \}$ that gives the 3-4-5 triangle but has infinitely many valid non-integer solutions that satisfy the triangle inequality.  Visually, the hypotenuse is sliding on the incircle.

[b]Follow-up:[/b]  How many triangles (not necessarily right) with integer sides have inradius $ 1$?  Inradius $ 2$?

[quote="t0rajir0u"] $ a \equal{} x \plus{} y, b \equal{} y \plus{} z, c \equal{} z \plus{} x$
[/quote]

Ravi's substitution? Very useful.

3,4,5 is indeed the only integer solution for any triangle. Using Heron's and a=x+y, b=y+z, c=x+z, we have
$ s \equal{} \sqrt {s(s \minus{} a)(s \minus{} b)(s \minus{} c)}$
$ s \equal{} (s \minus{} a)(s \minus{} b)(s \minus{} c)$
$ x \plus{} y \plus{} z \equal{} xyz$
If x is the highest of x,y,z, the LHS is $ \le 3x$, so $ yz\le 3$, from which we easily find 1,2,3 is the only integer solution.


Hello I am Dojo, a cellist and a pianist.
I have taught myself however, the violin, the viola and the bass.
In this thread, post your actual instrument(s).
Also the instrument you taught yourself.
I want to see if there is a correlation between actual and taught.

I play guzheng, piano, and percussion in band, none of which I taught myself, I have taught myself pieces though...

My actual instrument is trumpet and trombone.

I taught myself tuba, guitar, and keyboard.

My actual instruments are violin and piano.

I have taught myself the
viola
cello
bass
guitar
flute
daegum
oboe
basic drums/percussion

flute?
oboe?
Lol i wish i could touch one of those!!! (Ive seen them in my symphony but never touched one)
But violin or piano have almost no correlation.

I played them both in the school band, also filled in for percussion
several times...

Why? Are they supposed to have correlation..?  :maybe:


Find all functions $f: N_0 \to N_0$, (where $N_0$ is the set of all non-negative integers) such that $f(f(n))=f(n)+1$ for all $n \in N_0$ and the minimum of the set $\{ f(0), f(1), f(2) \cdots \}$ is $1$.

[hide]
Let x be the integer such that f(x) = 1.  Then f(f(x)) = f(1) = f(x)+1 = 2.
Similarly, f(2) = 3
f(3)=4
etc.
f(n) = n+1 for n > 0.
However, this means for n>0, f(n)>1.  However, we know that there is an integer x such that f(x)=1.  Since it can't be greater than zero and it must be non-negative, x=0 and f(0) = 1.

So f(n)=n+1 for all non-negative integers n.[/hide]

[hide]i think in order to get $f(f(f(n)))-f(f(n))=f(f(n))-f(n)$ it has to be linear.  Thus,  in order to have a minimum, it would have to be at 0 (because otherwise it wouldn't be linear) then it's $f(n)=n+1$.  Basically we have $f^{x+1}(n)=f(n)+x$ so it has to be $f(n)=n+1$.[/hide]


Prove that

2(a/b + b/c + c/a ) >= a/c + c/b + b/a + 3     where b+c>a>=b>=c>o

Since b + c > a, $c+a\geq c+b>b$ and $a+b\geq c+b>c$, the positive numbers a, b, c are the sidelengths of a triangle, and it remains to apply the inequality from http://www.mathlinks.ro/Forum/viewtopic.php?t=5842 .

  darij


quadratics
number theory
relatively prime
algebra
number theory unsolved

(xy)^2+(yz)^2+(zx)^2=z^4 with x,y,z are int

if $ (x,y,z)$ is a solution than $ (\minus{}x,y,z),(x,\minus{}y,z)...(\minus{}x,\minus{}y,\minus{}z)$ are also solutions so we can assume $ x,y,z$ are positive integers. (ruling out the solution $ (0,0,0)$)
we can assume without a loss of generality $ gcd(x,y,z)\equal{}1$ now denote $ gcd(x,y)\equal{}k$ consequently $ gcd(k,z)\equal{}1$ now putting $ x\equal{}ka$ and $ y\equal{}kb$ into equasion we get $ k^4(ab)^2\equal{}z^2(z^2\minus{}k^2(a^2\plus{}b^2))$ and since $ z^2(z^2\minus{}k^2(a^2\plus{}b^2))\equal{}z^4 \mod k$ and $ gcd(k,z)\equal{}1$ we can conclude that $ z^2(z^2\minus{}k^2(a^2\plus{}b^2))$ isn't devisable by $ k$ unless $ k\equal{}1$. therefore $ gcd(x,y)\equal{}1$

now denote $ A\equal{}z^2$
than $ A^2\minus{}A(x^2\plus{}y^2)\minus{}(xy)^2\equal{}0$ which means $ (x^2\plus{}y^2)^2\plus{}(2xy)^2$ is a perfect square
and because $ x,y$ are relatively prime $ (x^2\minus{}y^2,2xy,x^2\plus{}y^2)$ is a basic pitagorean triple and it is therefore imposible for $ (x^2\plus{}y^2)^2\plus{}(2xy)^2$ to be another perfect square. Therefore no solutions other than $ (0,0,0)$

i'm not compleately sure of the last conclusion. can someone give second opinion to it please?

Two comments:

1) The solution space is $ \mathbb{Z}\cdot\{(0,1,0),(1,0,0),(0,1,\pm 1),(1,0,\pm 1)\}$

2) As for the question, if $ (x^2\minus{}y^2,2xy,x^2\plus{}y^2)$ is a primitive Pythagorean triple with $ xy\not\equal{}0$, then the integer $ (2xy)^2\plus{}(x^2\plus{}y^2)^2$ cannot be a perfect square. In order to show this assume a minimal counterexample with regard to $ x^2\plus{}y^2$ and make the Pythagorean triple-[i]Ansatz[/i]: $ 2xy\equal{}2AB\,,\,x^2\plus{}y^2\equal{}A^2\minus{}B^2$, i.e. we can write $ A\equal{}A_1A_2,B\equal{}B_1B_2$ such that $ x\equal{}A_1B_1,y\equal{}A_2B_2$ holds. Substitution into the quadratic equation yields: $ A_1^2B_1^2\plus{}A_2^2B_2^2\plus{}B_1^2B_2^2\equal{}A_1^2A_2^2$. Division by $ A_1^2$ and remembering $ \gcd(x,y)\equal{}1$ yields the existence of some positive integer $ \lambda$ such that $ B_1^2\equal{}\lambda A_1^2\minus{}A_2^2$ is valid. Similarly, checking division by $ A_2^2$ we determine a positive integer $ \mu$ such that $ B_2^2\equal{}\mu A_2^2\minus{}A_1^2$ holds. Re-substitution yields $ A_1^2A_2^2(\lambda\mu\minus{}2)\equal{}0$, i.e. (i) $ \lambda\equal{}1,\mu\equal{}2$ or (ii) $ \mu \equal{}1,\lambda\equal{}2$. Both cases are similar, wlog we elaborate (i). This directly yields $ B_1^2\plus{}B_2^2\equal{}A_2^2\,,\,B_1^2\plus{}A_2^2\equal{}A_1^2$, i.e. $ (B_1,B_2,A_2)$ is a primitive Pythagorean triple with $ B_1^2\plus{}A_2^2$ aperfect square, contradicting the minimality of the counterexample, since $ A_2<x^2\plus{}y^2$.


algebra polynomial quadratics LaTeX trigonometry

solve $ 4z^{5}\plus{}z\equal{}0$ over $ C$

[hide]$ 4z^{5}\plus{}z \equal{} 0$
$ z(4z^{4}\plus{}1) \equal{} 0$, then either $ z \equal{} 0$ or $ 4z^{4}\plus{}1 \equal{} 0\Rightarrow z^{4}\equal{}\minus{}\frac{1}{4}\Rightarrow z \equal{}\frac{1}{2}\plus{}\frac{1}{2}i,\frac{1}{2}\minus{}\frac{1}{2}i,\minus{}\frac{1}{2}\plus{}\frac{1}{2}i,\minus{}\frac{1}{2}\minus{}\frac{1}{2}i$ [/hide]

[quote="bos1234"]How did you get the complex sonlutions?[/quote]
He took the fourth root of $ \minus{}\frac{1}{4}$. It yields those solutions. If you don't understand [i]why[/i] he got those solutions it's rather complex ( :rotfl: ) take the square root of $ \minus{}\frac{1}{4}$ (which are $ \frac{1}{2}i\text{ and }\minus{}\frac{1}{2}i$) and then take the square roots of those.

z^4 = -1/4,

z^2 = i*1/2, -i*1/2

z = sqrt(i*1/2), -sqrt(i*1/2), sqrt(-i*1/2), -sqrt(-i*1/2)

since sqrt(i) = sqrt(1/2)*(1+i), sqrt(-i) = i*sqrt(i) = sqrt(1/2)*(-1+i),

z = 1/2(1+i), -1/2(1+i), 1/2(-1+i), -1/2(-1+i)

Edit:

if you want to know how to get the square root of i, you can set a equation c^2= i, where c = a + b*i. This is one method, not the best, but the one I only know...

oh right i see how it works

kk thanks

I thought maybe we can use Demoivre's theorem

DeMoivre's theorem is a nice general way to look at problems of this type, but there is a [hide="fun elementary method"]$ 4z^{5}\plus{}z \equal{} z(1\plus{}4z^{4}) \equal{} z(1\plus{}2z\plus{}2z^{2})(1\minus{}2z\plus{}2z^{2})$

by the Sophie Germaine identity.  Then quadratic equation suffices.  :) [/hide]

[quote="timwu"]Ah, that one is much better:

if $ c^{2}\equal{} i$, then $ c \equal{} cos[\frac{(90\plus{}360k)}{2}]$ +$ i\sin[\frac{(90\plus{}360k)}{2}]$(k = 0,1) s o$ c \equal{} cos45\plus{}i\sin45$, $ \cos225\plus{}i\sin225$,

$ c \equal{}\pm\sqrt{\frac{1}{2}}(1\plus{}i)$.[/quote]

just fixing it up so i can read it  :lol:
use latex looks nicer

edit: How did you do this?

$ z^{2}\equal{}\frac{i}{2}$

$ \cos2\theta\equal{}0$
$ \sin2\theta$=$ \frac{1}{2}$

?

thanks

[quote="bos1234"]$ z^{2}\equal{}\frac{i}{2}$
$ \cos2\theta \equal{} 0$
$ \sin2\theta$=$ \frac{1}{2}$[/quote]
$ z$ is simple, $ z\equal{}\pm\frac{\sqrt{i}}{\sqrt{2}}$. If you need to expand this then use the identity $ \sqrt{i}\equal{}\pm\frac{\sqrt{2}}{2}\pm\frac{\sqrt{2}}{2}i$

For your trig questions use $ 2\theta\equal{}\phi$ then solve the equations for $ \phi$ in the usual way, and when you find $ \phi$ divide your answer(s) by two to obtain $ \theta$. Remember, if the question asks for all answers within the Unit Circle that is better to obtain the general solutions and divide those by two.


limit
factorial
function
logarithms
integration
calculus
calculus computations

This limit equalities:
$ \lim_{n\to\infty}\frac {{n!}^{\frac {1}{n}}}{n} = \lim_{n\to\infty}\(\frac {n!}{n^n})^{\frac {1}{n}} = \frac {1}{e}$

It doesn't work...if you wanna use Stolz

[hide]
The limit can be dealt with by putting bounds on the factorial function (even weak bounds should be okay since we're taking large roots of these factorials).

Express $ \ln n!$ as the sum $ \sum_{k\equal{}1}^n \ln k$, and note that $ \ln k\equal{}\int_k^{k\plus{}1} \ln k\ dx$, which we can approximate by the integral of $ \ln x$. Specifically, each term equals
\[ \int_k^{k\plus{}1} \ln x\ dx \minus{}\int_k^{k\plus{}1} \left( \ln x\minus{}\ln k\right)\ dx\]
Summing from $ k\equal{}1$ to $ n$,
\[ \sum_{k\equal{}1}^n \ln k\equal{}\int_1^{n\plus{}1}\ln x\ dx \minus{}\sum_{k\equal{}1}^n \int_k^{k\plus{}1} \left( \ln x\minus{}\ln k\right)\ dx\]
The big integral is equal to $ (n\plus{}1)\ln (n\plus{}1)\minus{}n$, which we can get by integrating by parts (differentiate $ \ln x$, integrate $ dx$).

Each of the small integrals are nonnegative, and the $ k$th one is at most $ \int_k^{k\plus{}1} \left( \ln (k\plus{}1)\minus{}\ln k\right)\ dx\equal{}\ln (k\plus{}1)\minus{}\ln k$ (since $ \ln x\leq \ln (k\plus{}1)$). It follows that the sum of the small integrals is between 0 and $ \ln (n\plus{}1)$. So we get
\[ n\ln (n\plus{}1)\minus{}n\leq \ln n!\leq (n\plus{}1)\ln (n\plus{}1)\minus{}n\]
This can be rewritten as
\[ \frac{(n\plus{}1)^n}{e^n}\leq n!\leq \frac{(n\plus{}1)^{n\plus{}1}}{e^n}\]
So let's use this on $ \sqrt[n\plus{}1]{(n\plus{}1)!}\minus{}\sqrt[n]{n!}$. An upper bound is
\[ \frac{(n\plus{}2)^\frac{n\plus{}2}{n\plus{}1}}{e}\minus{}\frac{n\plus{}1}{e}\]
and a lower bound is
\[ \frac{n\plus{}2}{e}\minus{}\frac{(n\plus{}1)^\frac{n\plus{}1}{n}}{e}\]
Both tend to $ \frac{1}{e}$, so we're done. (There are many ways to show $ n^{1/n}\rightarrow 1$ as $ n\rightarrow \infty$.)
[/hide]


After expanding and collecting like terms of the two polynomials below.

$f(x)=(1+x^{2}-x^{3})^{100}$, $g(x)=(1-x^{2}+x^{3})^{100}$

Which one has a larger coefficient for the $x^{50}$ term.

Extension: Find the two coefficients.

[hide]Because the x term is missing, these terms will always be alternating in sign when squared starting on the squared term.  However, the pattern of alternation is different.  In f(x), it's every other term.  In g(x), it's negative every third one starting on the $x^{2}$ coefficient.  Because of that, the 50th term will be positive in f(x) and negative in g(x).  Therefore, f(x) has the larger coefficient.
To prove this, just notice for f(x) that since 1 and $x^{2}$ are both positive, when taken to a power, all the even powers of x are even.  Also, since the coefficient on x is 0, the negative odd powers can only grow and stay negative by being multiplied against the positive $x^{2}$ term.  For g(x), one notices that the negative $x^{2}$ powers can only grow and stay negative when multiplied against the $x^{3}$ term.  This explains why every third power is negative starting with $x^{2}$.  Now, let's show all the coefficients of the powers 3n are positive when the leading power is 3n+3.  Although it's possible to have negative sums resulting from an odd number of multiplications of $-x^{2}$, we can show that a negative $x^{2}$ can generate $-x^{6n}$ where n is odd in less ways than $x^{3}$ can generate $x^{6n}$.  (If n is even, this is trivial.)
$-x^{2}$ can  generate $-x^{6}$ one way.
$x^{3}$ can generate $x^{6}$ three ways.
Now, the positives can be used by negative 2 when we generate the 18th power.  However, there must be an odd number of $-x^{6}$ in the multiplier.  This means that there's 3*3*1+1*1*1=10 ways to generate $-x^{18}$.
But there's 3*3*3+3*1*1=30 ways to generate $x^{18}$.
Notice that due to the initial disparity, the disparity grows wider.  This is because the formula for the positive is $3^{1}+3^{3}+3^{5}+...+3^{2n+1}$, while the formula for the negative is $3^{0}+3^{2}+...+3^{2n}$, meaning the positive will always be 3 times the negative.  A similar proof can be devised for the powers in the form 3n+1.[/hide]


Given that $ a,b,c,$ are real numbers such that $ a,b,c>0$ and $ a^2\plus{}b^2\plus{}c^2\equal{}1$

Prove:

$ a^{3/2}\plus{}b^{3/2}\plus{}c^{3/2}\geq3^{1/4}$

I brought another inequality down to this, but I'm not sure whether this is correct :blush:

[quote="ggbg90"]Given that $ a,b,c,$ are real numbers such that $ a,b,c > 0$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$

Prove:

$ a^{3/2} \plus{} b^{3/2} \plus{} c^{3/2}\geq3^{1/4}$

I brought another inequality down to this, but I'm not sure whether this is correct :blush:[/quote]

$ a \rightarrow 1$, $ b\equal{}c \rightarrow 0^{\plus{}}$.

Maybe you had a typo, it is just the opposite:
put $ a \equal{} \sqrt {x} \ b \equal{} \sqrt {y} \ c \equal{} \sqrt {z}$
Than we have $ x \plus{} y \plus{} z \equal{} 1$ and we will prove:
$ \sum_{cyc} x^{\frac {3}{4}}\leq 3^{\frac {1}{4}}$
It is true because it is equivalent to:
$ \sqrt [\frac {3}{4}]{\frac {x^{\frac{3}{4}} \plus{} y^{\frac{3}{4}} \plus{} z^{\frac{3}{4}}}{3}}\leq \frac {1}{3} \equal{} \frac {x \plus{} y \plus{} z}{3}$ which is true because:
$ \frac {3}{4} < 1$ :wink:

[quote="ggbg90"]aaah  - turns out I did have a typo. But how do you get the bit after "... it is equivalent to:"? Is it a known inequality?[/quote]

It is a well know inequality, the generalised power mean


1.Let there be given nine lattice points (points with integral coordinates)
in three dimensional Euclidean space. Show that there is a lattice point on the interior of one of the line segments joining two of these points. (Warning. It is easy to misread this problem. Take time to make sure you understand what is being asked for.)
2.To test your understanding, how many lattice points does one need in four dimensions to reach the same conclusion?

I want the answer for the 2. question as well.

Hint: PHP

what is the interior of a line segment?
 dya mean to say no three points must be collinear?

3 points can be collinear, but then the problem becomews trivial.
The interior of the line segment means that "it lies on the line segment".
Hope I hav clarified your doubts!

[quote]The interior of the line segment means that "it lies on the line segment". [/quote]

Sorry, I don't understand.  What's the difference between this statement and "three points being collinear"?[/quote]

Well, in the case of this question, we can have any lattice point lying on the segment connecting two of the given ones, it doesn't need to be one of the given lattice points.  Otherwise, this problem would be very trivial, as thealchemist said...

The answer for part 2 is indeed 17, and the solution for the 2 part is similar to the first.
Hope you all got my hint, use "pigeon hole principle"(or rather that's how I got it).


Let be $ n$ streets (horizontal) and $ m$ streets (vertical), if we can just go up and to the right, how many ways there exist to reach the corner $ (n,m)$ if we are in the "initial" corner $ (1,1)$

In each possible path there is n-1 "up" and m-1 "right".

Since there are m+n-2 steps in all, we consider the permutation (m+n-2)!. However, rearrange the n-1 "up" in the permutation gives same path, as well as the m-1 "right". Thus, total possible ways is $ \frac{(m\plus{}n\minus{}2)!}{(m\minus{}1)!(n\minus{}1)!}$


does anyone know the AMC's policy for ordering more tests?  i've recently had my teacher register for the A contest, without accurately gauging interest.  in the event that we have more people who want to take it, i'd like to know if it is possible to order more.  if so, what financial consequences and/or deadlines (if any) should i be aware of?

It is possible to order more.  The deadline is January 18th unless you want to pay extra shipping for expedited mailing.

There is an order form here:
[url]http://www.unl.edu/amc/d-publication/d1-pubarchive/2005-6pub/2006AMC1012/pdfs/11-061012TM-addbundles.pdf[/url]

See the [url=http://www.unl.edu/amc]AMC website[/url] for more information.


function
logarithms
number theory
relatively prime
algebra proposed
algebra

Characterize the multiplicative functions with real values such that $f(n+1)-f(n)$  tends to 0.

Do you mean there are some non-trivial examples?

It seems I have lost the vision of this problem... :(
I realized that my trivial examples includes two cases
a) $f(n)$ is an arbitrary multiplicative function s.t. $f(p^k)\to 0$ as $p^k\to\infty$;
b) $f(n)=n^a$, $a\in [0,1)$.
Are there other examples?

I think one can show that if $ f:\mathbb{N} \rightarrow
 \mathbb{R}_{+} $ is multiplicative , not bounded, such that the
 sequence $ f(n+1) - f(n) $ is bounded , then there exists
 $ \alpha \in (0,1] $ such that $ f(n) = n^{\alpha} $ for every $
 n \in \mathbb{N} $.

Myth, terribly sorry, I posted the wrong statement: the function must satisfy $f(xy)=f(x)+f(y)$ for relatively prime x and y and also $f(n+1)-f(n)$ tends to 0. Sorry very much again.

Do you mean $f(xy)=f(x)f(y)$? That is what I understand as a multiplicative function.

No, I want to say I posted the wrong statement in the begining. The corect formulation is : find functions f such that $f(xy)=f(x)+f(y)$ for relatively prime x,y and $f(n+1)-f(n)$ goes to 0 for large n. Again, sorry.  :(

In other words $f(x)$ is the logarithm of a multiplicative function.

I think we are to use ideas from http://www.mathlinks.ro/Forum/viewtopic.php?t=28867


calculus
integration
trigonometry
LaTeX
logarithms
derivative
calculus computations

I'm doing a problem set and I know I'll need help.

I'll post the problems and then my solutions as I get to them.
Help me get through them if you can

1. Evaluate the integrals: $ \int sin^{2}x cos^{4}x dx$
2. Evaluate the Integrals.
(a) $ \int {x^{m}}{(lnx)^{2}}\,dx$
(b) -i'm correct, so don't bother with this problem-
(c)$ \int \frac {1}{x^4\sqrt {a^2 \minus{} x^2} }\,dx$
thats the integral of 1/((x^4)((a^2-x^2)^(1/2))), but I can't seem to write that in latex [color=green][Make sure your {} balance. Fixed it for you. - Moderator][/color]
3. Calculate the Integrals.
(a) $ \int_0^{\frac {1}{2}} x*arcsin(2x)\,dx$
(b) $ \int_0^{\frac {\pi }{4}} {e^{3x}}{sin(4x)}\,dx$
(c) $ \int_1^2 {x^3}{lnx}\,dx$
4
do $ \int \sin(ln(x))\,dx$ with integration by parts

[quote](b) int((e^(ax))(cos(bx)),x)
>this one is a really long solution which basically boils down to integration by parts twice. I end up with (1/(a²+ b²))(e^(ax))(b sin(bx) + a cos(bx)). this seems right to me, so don't bother helping me with this. [/quote]

Let me show you a little bit of $ \text{\LaTeX}:$

Write the following:

\int e^{ax}\cos(bx)\,dx = \left(\frac{1}{a^2+b^2}\right)(b\sin(bx)+a\cos(bx))

only put a dollar sign at the beginning and a dollar sign at the end:

$ \int e^{ax}\cos(bx)\,dx = \left(\frac{1}{a^2+b^2}\right)(b\sin(bx)+a\cos(bx))$

Or you could write that right hand side as

\frac{b\sin(bx)+a\cos(bx)}{a^2+b^2} which becomes $ \frac{b\sin(bx)+a\cos(bx)}{a^2+b^2}.$

And yes, that is correct (OK, there should be a $ +C$ for the antiderivative).

If you use this way of writing the problems, you'll find that people here are more likely to respond to you.

Fix for [b]1. (c)[/b]:
[quote="onlyrevolutions"](c)$ \int \frac {1}{x^4\sqrt {a^2 \minus{} x^2}}\,dx$[/quote]
[hide="Hint for (4) if needed"]Let $ u \equal{} \ln x$. Then it should follow that $ e^u \equal{} x$[/hide]

4)
$ I\equal{}\displaystyle\int\sin(\ln{x})\,dx\equal{}\int(x)^{\prime}\sin(\ln{x})\,dx\equal{}x\,\sin(\ln{x})\minus{}\int{x}\left( {\sin(\ln{x})}\right)^{\prime}\,dx\equal{}
x\,\sin(\ln{x})\minus{}\int{x}\cos(\ln{x}){1\over{x}}\,dx\equal{}x\,\sin(\ln{x})\minus{}\int\cos(\ln{x})\,dx\equal{}x\,\sin(\ln{x})\minus{}J$.
Do the same for $ J$, then $ J\equal{}x\,\cos(\ln{x})\plus{}I$. So $ 2\,I\equal{}x\,\sin(\ln{x})\minus{}x\,\cos(\ln{x})$.

[hide="Number 2.c)"]$ \int \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}}\,dx \equal{} \lambda \frac {\sqrt {a^2 \minus{} x^2}}{x^n} \plus{} f(x)$.

Now differentiate both sides to get

$ \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}} \equal{} \minus{} \lambda \frac {x^{n \plus{} 1} \plus{} n(a^2 \minus{} x^2)x^{n \minus{} 1}}{x^{2n} \sqrt {a^2 \minus{} x^2}} \plus{} f'(x)$

$ \equal{} \minus{} \lambda \frac {x^{n \plus{} 1} \minus{} nx^{n \plus{} 1}}{x^{2n} \sqrt {a^2 \minus{} x^2}} \minus{} \lambda n a^2 \frac {x^{n \minus{} 1}}{x^{2n} \sqrt {a^2 \minus{} x^2}} \plus{} f'(x)$

$ \equal{} > \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}} \equal{} \lambda (n \minus{} 1) \frac {1}{x^{n \minus{} 1} \sqrt {a^2 \minus{} x^2}} \minus{} \lambda n a^2 \frac {1}{x^{n \plus{} 1} \sqrt {a^2 \minus{} x^2}} \plus{} f'(x)$

Now choose $ n \equal{} 3$ and so $ \lambda \equal{} \minus{} \frac {1}{3a^2}$, and we get

$ \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}} \equal{} \minus{} \frac {2}{3a^2} \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}} \plus{} \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}} \plus{} f'(x)$

which implies $ f'(x) \equal{} \frac {2}{3a^2} \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}} \equal{} > f(x) \equal{} \frac {2}{3a^2} \int \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}}\,dx$.

In a similar fashion, we have

$ \int \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}}\,dx \equal{} \alpha \frac {\sqrt {a^2 \minus{} x^2}}{x^k} \plus{} g(x)$

$ \equal{} > \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}} \equal{} \alpha (k \minus{} 1) \frac {1}{x^{k \minus{} 1} \sqrt {a^2 \minus{} x^2}} \minus{} \alpha k a^2 \frac {1}{x^{k \plus{} 1} \sqrt {a^2 \minus{} x^2}} \plus{} g'(x)$

and now we must choose $ k \equal{} 1$ and so $ \alpha \equal{} \minus{} \frac {1}{a^2}$, which leads us to

$ \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}} \equal{} \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}} \plus{} g'(x)$

$ \equal{} > g'(x) \equal{} 0$

$ g(x) \equal{} C$.

Therefore, we have $ \int \frac {1}{x^2 \sqrt {a^2 \minus{} x^2}}\,dx \equal{} \minus{} \frac {\sqrt {a^2 \minus{} x^2}}{a^2 x} \plus{} C$, and so $ f(x) \equal{} \minus{} \frac {2 \sqrt {a^2 \minus{} x^2}}{3a^4x} \plus{} C'$

and finaly

$ \int \frac {1}{x^4 \sqrt {a^2 \minus{} x^2}}\,dx \equal{} \minus{} \frac {\sqrt {a^2 \minus{} x^2}}{3a^2x^3} \minus{} \frac {2 \sqrt {a^2 \minus{} x^2}}{3a^4x} \plus{} C'$.

Observe that, in this way, we have transformed an integration problem into a differentiation problem.[/hide]

[hide="Number 1"]$ \int \sin^2x \cos^4 x\,dx \equal{} \int \sin^2x \cos^2x \cos^2x\,dx \equal{} \int \left[\frac{\sin(2x)}{2}\right]^2 \cdot \frac12 [1 \plus{} \cos(2x)]\,dx$

$ \equal{} \frac18 \int [\sin^2(2x) \plus{} \sin^2(2x) \cdot \cos(2x)]\,dx \equal{} \frac18 \int \frac12 [1 \minus{} \cos(4x)]\,dx \plus{}$

$ \plus{} \frac{1}{16}\int \sin^2(2x) \cdot 2 \cos(2x)\,dx \equal{} \frac{x}{16} \minus{} \frac{\sin(4x)}{64} \plus{} \frac{\sin^3(2x)}{48} \plus{} C$.[/hide]

[hide="Number 2.a)"]Making the substitution ln x = u we get

$ \int x^m (\ln x)^2\,dx \equal{} \int u^2 e^{u(m\plus{}1)}\,du \equal{} (Au^2 \plus{} Bu \plus{} C) e^{u(m\plus{}1)}$.

By differentiation we get

$ u^2 e^{u(m\plus{}1)} \equal{} [A(m\plus{}1) u^2 \plus{} (Bm \plus{} B \plus{} 2A) u \plus{} B \plus{} C(m\plus{}1)] e^{u(m\plus{}1)}$

and so $ A \equal{} \frac{1}{m\plus{}1}$, $ B \equal{} \minus{} \frac{2}{(m\plus{}1)^2}$, $ C \equal{} \frac{2}{(m\plus{}1)^3}$. Therefore,

$ \int x^m (\ln x)^2\,dx \equal{} \left[\frac{(\ln x)^2}{m\plus{}1} \minus{} \frac{2 \ln x}{(m\plus{}1)^2} \plus{} \frac{2}{(m\plus{}1)^3}\right] \cdot x^{m\plus{}1} \plus{} C$.

Of course, this formula is valid only for $ m \neq \minus{}1$. For m = -1 we have

$ \int \frac{(\ln x)^2}{x}\,dx \equal{} \frac{(\ln x)^3}{3} \plus{} C$.[/hide]

[hide="Number 4"]$ \int \sin(\ln x)\,dx \equal{} \Im \left[\int e^{i \ln x}\,dx\right] \equal{} \Im \left[\int x^i\,dx\right] \equal{} \Im \left[\frac{x^{1 \plus{} i}}{1\plus{}i} \plus{} C \plus{} iC'\right]$

$ \equal{} \Im \left[\frac{x \cdot e^{i \ln x}(1\minus{}i)}{2} \plus{} C \plus{} iC'\right] \equal{} \Im \left[\frac{[x \cos(\ln x) \plus{} ix \sin(\ln x)](1\minus{}i)}{2} \plus{} C \plus{} iC'\right]$

$ \Im \left[\frac{x \cos(\ln x) \minus{} ix \cos(\ln x) \plus{} ix \sin(\ln x) \plus{} x \sin(\ln x)}{2} \plus{} C \plus{} iC'\right]$

$ \equal{} \frac12 [x \sin(\ln x) \minus{} x \cos(\ln x)] \plus{} C'$.

Observe that we have also concluded that

$ \int \cos(\ln x)\,dx \equal{} \frac12 [x \sin(\ln x) \plus{} x \cos(\ln x)] \plus{} C$.[/hide]

[hide="Number 3.a)"]Making 2x = sin(u) we have

$ I \equal{} \int_0^{1/2}x \arcsin(2x)\,dx \equal{} \frac14 \int_0^{\pi/4}u \sin u \cos u\,du$

$ \equal{} \minus{} \frac {1}{16} \int_0^{\pi/4}u \cdot [ \minus{} 2 \sin(2u)]\,du \equal{} \minus{} \frac {1}{16} \left[[u \cdot \cos(2u)]_0^{\pi/4} \minus{} \int_0^{\pi/4} \cos(2u)\,du\right]$

$ \frac {1}{32} \int_0^{\pi/4}2 \cos(2u)\,du \equal{} \frac {1}{32} [\sin(2u)]_0^{\pi/4} \equal{} \frac {1}{32}$.[/hide]

[hide="Number 3.c)"]$ \int_1^2 x^3 \ln x\,dx \equal{} \left[\frac14 x^4 \ln x\right]_1^2 \minus{} \frac14 \int x^4 \cdot \frac1x \,dx \equal{} 4 \ln 2 \minus{} \frac14 \left[\frac{x^4}{4}\right]_1^2 \equal{} 4 \ln 2 \minus{} \frac{15}{16}$.[/hide]

2.(a)

Let $ \ln x \equal{} t$, then $ dx \equal{} e^t\ dt$.

$ \therefore \int x^m(\ln x)^2\ dx \equal{} \int t^2e^{(m \plus{} 1)t}\ dt$.

Answer:

For $ m \equal{} \minus{} 1$, we have $ \frac {1}{3}t^3 \plus{} C$.

For $ m\neq \minus{} 1$, we have $ \frac {1}{m \plus{} 1}t^2e^{(m \plus{} 1)t} \minus{} \frac {2}{(m \plus{} 1)^2}te^{(m \plus{} 1)t} \plus{} \frac {2}{(m \plus{} 1)^2}e^{(m \plus{} 1)t} \plus{} C$.

4. Just put $ \ln x \equal{} t$.

Or

$ (x\sin \ \ln x)'\equal{}\sin \ \ln x\plus{}x\cos \ ln x\cdot \frac{1}{x}$.

$ (x\cos \ \ln x)'\equal{}\cos \ \ln x\minus{}x\sin \ \ln x\cdot \frac{1}{x}$.

$ \therefore 2\sin \ \ln x \equal{}(x\sin \ \ln x)'\minus{}(x\cos \ \ln x)'$, yielding the answer.

[quote="onlyrevolutions"]2 C and 3 B still don't make sense to me.[/quote]

$ (e^{3x}\sin 4x)'\equal{}....$

$ (e^{3x}\cos 4x)'\equal{}...$

For 2.c) you can also make the substitution $ x \equal{} a \sin u$:

$ P \equal{} \int \frac{a \cos u}{a^4 \sin^4 u \sqrt{a^2 \minus{} a^2 \sin^2u}}\,du \equal{} \frac{1}{a^4} \int \frac{1}{\sin^4 u}\,du$.

Can you proceed from here?


Why do I think they might be?

Firstly, it shows interest in reading AND writing, both of PARAMOUNT impotance in any area of college.

Secondly, it shows a wanting to contribute to the community

Thirdly, they're easily verifiable.

Perhaps these would make Amazon.com and Wikipedia (by the way, I mean writing articles for 'em) beneficial for college apps as an EC? I would much prefer writing reviews on Amazon than to go outside and to play a sport that I'll never master.

Thanks,

Simfish

Well, "follow your passions" is standard advice for choosing and applying to colleges, so I wouldn't tell you not to do this, nor would I tell you not to mention it on an application. I don't know--AT ALL--how an admissions office might interpret that interest of yours.

What do some of you recent admittees think?

That is a very interesting question, and one I'm sure is about to spark up a fierce debate  :).  How many articles have you written?

I think the Wiki encyclopedia might be a good idea, only if there is some way to proove you've written all of those articles.

But the Amazon idea, really isn't worth putting down I guess for me.  Anyone can right an amazon review, it doesn't need to be good or anything.

It's easy to check whether someone has written a Wikipedia article. (You can check that I have written a few if you want to.) But why should it help on a college application? Really, why do people get so worked up about what will help on a college application and what won't? Do what you want to do! Of course, very few people (only the admissions people, in fact) know what helps on a college application, and it is unlikely that they will tell anyone else since that would mess up the admissions process.

Hmmm... This might be a problem.... Anyone can do certain community service too... As for me, I've actually just started on both. I started reviewing on Amazon last month (3 reviews so far; 10 helpful votes) and Wikipedia a couple of days ago (but went on a Wikirampage).

What year are you?  If you're a senior, it will seem like (and may be... :) ) a desperate attempt to add another thing to your resume.

This is certainly a creative idea, but I don't think colleges would really appreciate it.  Perhaps if you make your own website and review books or write an online encyclopedia (perhaps a specialized one - on math, maybe?), that would stand out.  But really, lots of people write Amazon reviews, and there's no way of being sure that you actually wrote the reviews.  It could have, say, been your mom.

I'm currently a sophomore so I have plenty of time.

Oh, I see... Designing a website would stand out?? I've already designed three! (all before I was even an eighth grader) But many teens design their own webpages too.  And would admissions officers have the time to look at a website?

[quote="mathfanatic"] But really, lots of people write Amazon reviews, and there's no way of being sure that you actually wrote the reviews.  It could have, say, been your mom.[/quote]

Truly this is a problem that spans many activities (though it's usually hard to get your mom to fit into your football uniform and play for you).  It's a credibility problem, but one that is often mitigated by a consistent set of activities and performances.

I don't see why writing for Wikipedia or Amazon would really help your college app.  If you see fit you should write it (it cant hurt you) but don't both sites
allow anybody to write articles without checking them?

Amazon.com is evil.  Wikipedia is not.

I do write for Wikipedia quite a bit; I wrote the entire original draft of the article [url=http://en.wikipedia.org/wiki/Integer_square_root][i]Integer square root[/i][/url], in fact.  And I've contributed in random other places as well, including Wikisource and Wikibooks (where I've taken it upon myself to [i]try[/i] and revamp the [url=http://en.wikibooks.org/wiki/Calculus]Calculs Wikibook[/url]).

I think it would be great if contributing to wikis were beneficial to college applications, but I don't see how they could be; too many factors need to be included, and you could simply have somebody to the work for you.

-iamunknown73


Alcumus FTW MATHCOUNTS probability geometry AMC AIME

First off, if this is the wrong place for this post, I apologize. I am new to the forum.

Okay. I just recently found this site and decided to give it a shot. I got a keg (if the analogy makes any sense). Not only does AoPS have a forum but it also has Alcumus, online courses, AND a series of acclaimed books. This is all slightly intimidating. My question is where do I begin? I more or less missed out on all of the competition classes. Should I check out one of the other classroom courses? And what about the books and Alcumus? In other words, how should one proceed in the pursuit of problem-solving proficiency?

I do realize that the posts in this section have to be applicable to as many people as possible. I am probably disregarding this requirement when I mention that I am an eleventh grader who got a mid-60 on the AMC 12 last year . . . .

[quote="PoorPythagoras"]I am probably disregarding this requirement when I mention that I am an eleventh grader who got a mid-60 on the AMC 12 last year . . . .[/quote]
That is actually very good for starting you off. For these types of questions, they might better belong in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url], but that's okay.

AoPS also has [url=http://www.artofproblemsolving.com/Edutainment/g1/index.php]FTW[/url] (For The Win) which is a MathCounts Countdown type of competition, with fast-paced head-to-head competing. I think this should only be used to increase speed, rather than for practice.

Alcumus is mainly for introductory counting and probability, but it is a great review. It does cover the counting and probability through most of the AMC 12, so it would be good resource. This, the Intro to Counting and Probability book, and Intro to C+P class pretty much cover the same thing, but in different ways. A combination of these should probably be chosen based on your learning style.

The classes and books kind of go hand-in-hand. Basically, every person learns different ways, so they need to choose the right one for them. This is different for individuals, so there is no general advice. Pretty much every book has a class, so you should do whichever you learn best through, or both. The full curriculum goes through Introduction to Algebra, Number Theory, Counting and Probability, and Geometry, then the Intermediate of most of those. Intermediate Number Theory doesn't have a book and it has a seminar (like class), and there is a bookless Olympiad Geometry class. This covers stuff through AIME and some USAMO.

Finally, the most important thing is to practice using old AMCs or other problem-solving competitions. AoPS has a bunch of them in the [url=http://www.artofproblemsolving.com/Forum/resources.php?c=182]Resources Section[/url] and the [url=http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions]AMC Test section[/url] of its Wiki.

What I think you should do is start by taking various pre-tests and post-tests of the classes and reading through table of contents and excerpts of the books to determine your level. You can use Alcumus to patch up on counting and probability, and use the practice problems to make sure you know it. Then, by books and take classes appropriately. Finally, make sure you practice a lot. If you get a question wrong, make sure you know what you did wrong and to avoid that mistake the next time. Once you feel that you can get most of the questions right, start timing yourself like the real test (75 minutes) and start learning how to distribute your time.

Good luck!

[quote="PoorPythagoras"]Okay. I just recently found this site and decided to give it a shot. [/quote] To start out, I would look around the Middle School Contest section and try to solve problems posted there.
http://www.artofproblemsolving.com/Forum/index.php?f=532 is good too. If you would like to start learning a specific topic, you can buy a book from somewhere or use alcumus. There are also several online databases with a multitude of questions. Lastly, you can try practice tests, starting on things such as Mathcounts/AMC 8, and slowly getting harder to AMC 10, 12, and then AIME.

So if I were you, I would do this:
Lurk on Middle School forums I showed you and do problems (if too easy go to highschool)
Then, try some practice problems of AMC 8 and Mathcounts, work your way up to AMC 10.
Either buy a book or try to study more of the subjects you are worse in, especially try to get good at discrete math.
After, take more AMC 10/12 problems.

I'm in a similar situation. I opted for the books. The problem is that the going is incredibly slow-not because I don't understand the material but because I ask a LOT of questions about the material. For example, I am reading through the introductory number theory book right now. I only read a page and a half so far and I'm already asking how to relate the process of counting to numbers to operations THROUGH A NEW NOTATION. Am I just being stupid? Is too much curiosity a bad thing?

[quote="OnASlope"] Am I just being stupid? Is too much curiosity a bad thing?[/quote]
Not stupid, but too impatient. Read the full chapter first and then see what questions make sense and still remain unanswered ;). Curiosity is never bad but it should be timed properly. I could also rush to google and start searching for the etimology of the English pronoun "I" after reading the first three symbols of your post, but would there be any chance to ever get to the meaning of it then?


If $ a,b,c$ are the sides of triangle and $ x,y,z$ are real numbers then prove:
$ (x\plus{}y\plus{}z)(\frac{xc^2}{a^2}\plus{}\frac{ya^2}{b^2}\plus{}\frac{zb^2}{c^2}) \ge (\frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2})(a^2yz\plus{}b^2zx\plus{}c^2xy)$.

[quote="soruz"]If $ a,b,c$ are the sides of triangle and $ x,y,z$ are real numbers then prove:
$ (x \plus{} y \plus{} z)(\frac {xc^2}{a^2} \plus{} \frac {ya^2}{b^2} \plus{} \frac {zb^2}{c^2}) \ge (\frac {1}{a^2} \plus{} \frac {1}{b^2} \plus{} \frac {1}{c^2})(a^2yz \plus{} b^2zx \plus{} c^2xy)$.[/quote]
From $ (x\vec{MA} \plus{} y\vec{MB} \plus{} z\vec{MC})^2 \geq 0$
We have
$ (x \plus{} y \plus{} z)(xMA^2 \plus{} yMB^2 \plus{} zMC^2) \geq a^2yz \plus{} b^2zx \plus{} c^2xy$
We can choose $ M$ such that $ MA^2 \equal{} \frac {c^4b^2}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$

$ MB^2 \equal{} \frac {a^4c^2}{a^2b^2 \plus{} b^2 c^2 \plus{} c^2a^2}$
$ MC^2 \equal{} \frac {b^4a^2}{a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2}$
 :oops:


superior algebra
superior algebra unsolved

Let $r,s,t$ be integer greater than 1.prove that there exists finite group $G$ that contains $a$ and $b$ with order $s$ and $t$ respectively and $ord(ab)=r$

Any idea for this (indeed) cute result ? I haven't given much thinking yet but it seems lik esomehow we'll have to play with noncommutative groups ...

$G=\left\langle a,b ~|~ a^s = 1 , b^t = 1 , (ab)^r = 1 \right\rangle$   :roll:

There is probably some math I don't know in order to understand your answer ... in other words, how do I know the group you describe actually exists ?

http://en.wikipedia.org/wiki/Presentation_of_a_group

It's just a guess :)

-oo-, that group of yours isn't necessarily finite (in fact, it's infinite except for a few particular cases). In the problem posted by sam-n, the group is required to be finite.

Edit:

Julien, I've edited the message. I was talking to -oo-, merely trying to point out that his construction does not work for this problem. My post had nothing to do with what you do or do not know about group presentations :).

No, it's just that I don't know anything about group presentations. Thus in the answer given by oo, I am not able to understand what makes the given presentation valid (i.e, I would understand why $<a^{m}=1,b^{n}=1>$ exists (we take a free group), but not why $<a^{m}=1,b^{n}=1,(ab)^{r}=1>$ is well defined). Anyhow, it's a good occasion to read some intro to group presentations  :D.  Thanks for your answers !


pigeonhole principle
combinatorics proposed
combinatorics

Let  $F=\{1,2,...,100\}$ and let $G$ be any $10$-element subset of $F$. Prove that there exist two disjoint nonempty subsets $S$ and $T$ of $G$ with the same sum of elements.

Consider all subsets of G. There are $2^{10}-1$. There are $2^{10}-1$ sums, but all sums have only$\leq 100+99+98+97+.+91<1000$possible values and so, by pigeonhole theorem there are two equal sums. Since there are two subsets with the same sum of elements, let them be A and B, then $A - B$ si $B-A$ are disjoint subsets of G with same sum of elements.

See here also: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=48464


The five-digit number $ 6a47b$ is divisible by 18. What is the smallest possible value of $ a\minus{}b$?

$ 6+a+4+7+b \equiv a+b-1 \pmod{9}$ and $ b$ is even integer.

If $ b=0$, $ a=1$.
If $ b=2$, $ a=8$.
If $ b=4$, $ a=6$.
If $ b=6$, $ a=4$.
If $ b=8$, $ a=2$.

Thus, smallest possible value of $ a-b$ is $ 2-8=\boxed{-6}$.


Let $a > 1$ and $r > 1$ be real numbers.
(a) Prove that if $f : \mathbb{R}^{+}\to\mathbb{ R}^{+}$ is a function satisfying the conditions
(i) $f (x)^{2}\leq ax^{r}f (\frac{x}{a})$ for all $x > 0$,
(ii) $f (x) < 2^{2000}$ for all $x < \frac{1}{2^{2000}}$,
then $f (x) \leq x^{r}a^{1-r}$ for all $x > 0$.
(b) Construct a function $f : \mathbb{R}^{+}\to\mathbb{ R}^{+}$ satisfying condition (i) such that for all $x > 0, f (x) > x^{r}a^{1-r}$ .

[quote="arkhammedos"]:maybe: i think $r$ shoud be rational[/quote]
Why you think so?

[quote]Why you think so?[/quote]
because they say there is no reel exponent (i am yet in high school)
we learned just rational  $\frac{p}{q}$
-qestion is this ecriture existe $x^{\pi}$. :maybe:


limit
calculus
derivative
geometric series
algebra
binomial theorem

Hi there,

I was reading a book the other day, and one of the questions in it was the following:

what is the sum of the following series:

1 + (1/4) + ((1*4)/(4*8))+((1*4*7)/(4*8*12)).....

I tried everything I could on the question,  and from just simplyfying the terms, and resolving them into partial terms, i could see that it did converge.

I was wondering if anyone had some hints or tips, as to what this infinite series adds up to, and perhaps also as to the kind of series that it is.

Thank you

Sincerely JHM

Your series is

$S = 1+\sum_{n=1}^{\infty}\frac{1 \cdot 4 \cdot 7 \cdot ... \cdot (3n-2)}{4 \cdot 8 \cdot 12 \cdot ... \cdot (4n)}$.

We have that $\frac{a_{n+1}}{a_{n}}= \frac{3n+1}{4n+4}$ and so the series converges by the D'Alembert Test because

$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \frac{3}{4}< 1$.

Actually, lambern314 is looking for the sum of the series...
He already knows it converges  :maybe:

On my TI84, I made a program that calculated the sum to be $4.416402969$...
Anyone recognize that?

Hey, this is me again,

doing this on my calculator for the first 3 terms, gives 1+(1/4)+(1/8)+(7/96).....
the partial sum is 1.44791666, then 1.493489583, then 1.523111979, then 1.542860243, then  1.556260851, playing around my calculator, this number seems to be a little higher that the cubed root of three (1.44224957), and quite close to the cubed root of four (1.587401052). I dont know if this helps.

Thank you for your time

What kind of book was it?  It would help if we knew roughly what level of techniques are required.

It was an advanced algebra book, the excersice was in the back, in the additional excersices section. covers precal with more topics.

The exact sum is definitely a calculus topic, which doesn't belong here.

[hide="Solution"]Let $f(x)=1+\frac14x+\frac{1\cdot 4}{4\cdot 8}x^{2}+\cdots=\sum_{n=0}^{\infty}a_{n}x^{n}$ for $x$ near zero. We want to calculate $f(1)$.
We have $a_{n}=\frac{3n-2}{4n}a_{n-1}$, or $(3n-2)a_{n-1}=4na_{n}$.

Now, $x^{\frac13}f(x)=\sum_{n=0}^{\infty}a_{n-1}x^{n-\frac23}$ and $\frac{d}{dx}\left(x^{\frac13}f(x)\right)= \sum_{n=1}^{\infty}x^{n-\frac53}\frac{3n-2}{3}a_{n-1}.$ Also, $f'(x)=\sum_{n=1}^{\infty}\frac{4n}{4}a_{n}x^{n-1}.$

By the coefficient relation, $4f'(x)=3x^{\frac23}\frac{d}{dx}\left(x^{\frac13}f(x)\right)=3xf'(x)+f(x)$ and $f'(x)=\frac1{4-3x}f(x)$
That gives $f(x)=C\exp\left(-\frac13\ln(4-3x)\right)=\frac{C}{\sqrt[3]{4-3x}}$ for some real $C$. Since $f(0)=0$, $f(x)=\frac1{\sqrt[3]{1-\frac34x}}$, and $f(1)=\sqrt[3]{4}$.[/hide]

Thank you jmerry, the explanation is kind of difficult, but looks as if it makes sense. But I think there might be a more easy solution, without calculus, because my algebra book definitely does not cover derivatives  or anything about calculus. I'm going to read into it alot more and see if I can find something similar to this. Is this a special type of geometric series?

Thanks again

It's the binomial series- Newton's favorite tool.

Also, limits and convergence are calculus. If there's an infinite sum- that's a limit. This particular series is complicated enough that I don't see any way to do it short of Taylor series; any method that gives you the sum will also give you $f(x)$ for any $x$ at which the series converges. That material is traditionally taught at the end of the first year of calculus, and a typical calculus student would have a very hard time solving this problem.

FWIW:

Things are easy when the terms in denominator goes  as product of  (a,2a,3a,4a) and numerator has products of an arithmetic series.

[hide]If one knows the   expansion of $(1+x)^{n}$ as  $1+nx+\frac{n(n-1)}{2}x^{2}+\dots$
One  may be familiar with  the case when both x and n are negative, you get all positive terms,  eg:

$(1-x)^{-n}= 1+n x+\frac{n(n+1)}{2!}x^{2}+\frac{n(n+1))(n+2)}{3!}x^{3}+\dots$

Keeping that in mine, when one looks at the series above,   one can't help but notice that the denominator   goes as $1! , 2!, 3!$  as long as you divide by multiples of 4 . And the common difference  in terms in the numerator is 3 ...[b] Hint Hint .. take out (3/4) [/b]

 IOW  The series could be looked as

$1+\frac{(1/4)}{1!}+\frac{(1/4)(4/4)}{2!}+\frac{(1/4)(4/4)(7/4)}{3!}+\dots$

So looking at  expansion of (1-x)^(-n), you do notice that  a factor of (3/4) will do the trick,  that is, looking at the[b] third [/b]term the numerator  above could be thought as

$(3/4)^{3}(1/3)(4/3)(7/3)$ or as $(3/4)^{3}(1/3)(1/3+1)(1/3+2)$ etc..

That way  the answer would be  $(1-3/4)^{(-1/3)}$
(In some ways    the answer/logic is same as  jmerry but perhaps one can understand it by comparing with Binomial Theorem)
In fact  this way, it is easy to see that:
$S = 1+\sum_{n=1}^{\infty}\frac{a (a+d)(a+2d)(a+3d) \cdots (a+(n-1)d))}{b \cdot 2b \cdot 3b \dots nb}= (1-(d/b))^{-(a/d)}$  [/hide]

wow.. :fool: those are nice solutions to a nice problem.. thank you Gyan and jmerry! :omighty:

Thank you very much everyone that helped, that has put my mind to rest for now.

SIncerely

JHM


Given $ a,b,c \ge 0$.Prove that:
$ a^3\plus{}b^3\plus{}c^3 \plus{}3abc \ge \sqrt{\frac{2}{a^2\plus{}b^2\plus{}c^2}} [a^3(b\plus{}c)\plus{}b^3(c\plus{}a)\plus{}c^3(a\plus{}b)] \ge ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)$

[quote="babyloverain"]Given $ a,b,c \ge 0$.Prove that:
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 3abc \ge \sqrt {\frac {2}{a^2 \plus{} b^2 \plus{} c^2}} [a^3(b \plus{} c) \plus{} b^3(c \plus{} a) \plus{} c^3(a \plus{} b)] \ge ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)$[/quote]
Try a=b=c=1  :wink:


The sum of the following seven numbers is exactly 19:
\[a_1=2.56,\qquad a_2=2.61,\qquad a_3=2.65,\qquad a_4=2.71,\]
\[a_5=2.79,\qquad a_6=2.82,\qquad a_7=2.86.\]
It is desired to replace each $a_i$ by an integer approximation $A_i$, $1 \le i \le 7$, so that the sum of the $A_i$'s is also 19 and so that $M$, the maximum of the "errors" $|A_i - a_i|$, is as small as possible.  For this minimum $M$, what is $100M$?

[hide="Answer"]To sum to $19$, we need two 2's and five 3's. Obviously, rounding any number greater than one of the first two to $2$ will give a greater error than rounding the first two to $2$, and the other five will round to $3$. The largest error is therefore $|A_2-a_2|=0.61=M$, so $100M=\boxed{061}$.[/hide]

[quote="hockeyadi23"][quote="me@home"]Hey Billy, how do you post AIME problems?
Also, how did you do on the 2006 AIME??[/quote]

Yeah how do you type in the source?[/quote]

Ask Valentin; he lets one be able to post the source, or not be able to, as the case may be.

[quote="qweretyq"]Is it just me or is $0.61 * 100 = 61$ not $610$?[/quote]

Naaah  :D

[quote="calc rulz"][quote="qweretyq"]Is it just me or is $0.61 * 100 = 61$ not $610$?[/quote]

It's just you  :P[/quote]

I fixed it for Jesusfreak ;)

[quote="cdymdcool"]i don't think the first seven numbers add up to 19.... :([/quote]

He's right, it adds up to 18.99  :P

Really though, it does.

[quote="Yongyi781"]That issue is solved immediately by letting $ a_7 \equal{} 2.87$ instead of $ 2.86$.[/quote]

Yes but it says a_7=2.86

We must sue them!! How could they make this huge huge error?!

 :P  :P  :P

I don't think the laws of mathematics apply in court.

Back on topic, from other sources, in the 1985 AIME $ a_6$ was $ 2.82$, not $ 2.81$.

Problem statement fixed. Good catch, guys.

p12: We know that we want to be using two's and three's since that's what these numbers round up and down to. There are a total of seven numbers, so the only way this can work is with $5$ three's and $2$ two's. To minimize, we'd want to assign $3$ to the higher numbers and $2$ to the smallest two - so the maximum in that scenario would be $0.61 \implies \boxed{61}$.


Fie $A \in M_3 (C)$ cu proprietatea ca $detA=TrA=0.$
Sa se calculeze $\mathop {\lim }\limits_{n \to \infty } tr(A^{2n + 1} )$

Cayley-Hamilton gives

$A^3-tr(A)A^2+tr(A^{*})A-det(A)I_3=0$ where $A.A^{*}=det(A)I_3$

we have $det(A)=tr(A)=0$

$A^3=kA$ by induction  $A^{2n+1}=k_n.A$ this gives $Tr(A^{2n+1})=0$


Would anyone suggest a good book for practicing for the AIME exam?

Well, the best way to go is really practice past exams; you can find them for free online.
AoPS Vol. II helps, if you want a book, although I don't think it'll do you much good this close to the exam.


[quote="mathcrazed"]lol thats seriously eg?[/quote]

why woudlnt taht be him.


Hello,

Does anyone know how to find an equivalent distribution for $ \delta(t^{2007} \plus{} 1)$ ?
I need to know the answer for two cases, a) t real number, b) t complex number
I found the formula $ \delta(g(x)) \equal{} \sum_{k\equal{}1}^n\frac{\delta(x \minus{} x_k)} {|g'(x_k)|}$, but I don't know what happens when $ x$ is real and when $ x_k$ is a multiple root of $ g(x)$.
$ t$ should be the time in signal processing applications, so complex time doesn't make too much sense.

Thank You

At multiple roots, there fails to be a distribution at all- you get $ \pm\infty$ when applied to any test function nonzero at that point. Outside $ \mathbb{R}$, you need new definitions to make any sense of things- after all, the standard test function space is a space of functions on the real line. I don't think it's worth bothering with at all.


Everyone in a class of 35 students takes math and history.
8 students received an A in history, 15 received an A in
math and 3 received an A in both courses. How many
students did not receive an A in either course?

To find out the number of A's, add the number of A's in each class and subtract the doubles:
8 + 15 = 23
23 - 3 = 20.
There are 35 people in the class. therefore, the number of people who did [b]not[/b] get A's = 35 - 20 = [u]15[/u].


The BITSAT 2009 Official Sample paper (application) has been released and is available for download on the home-page.
Or click [url=http://www.bitsadmission.com/bitsat/BITSAT-2009.exe][img]http://www.bitsadmission.com/images/bullet.gif[/img] here[/url] to start download.

-Best of luck!

Also post your BITSAT centres and timings. It is good to get acquainted with people of your exam batch so that you are comfortable in the exam hall.

Mine is:
$ 3^{rd}$ [b]June, 2009
Mumbai,
4:30pm[/b]

Hey there are so many appearing for BITSAT. You lied!  :furious:

I guess there was no great effort involved as such, huh? I am very [b]innocent[/b], you see.

yes
that's y u were befooled by a fool :rotfl:

may 15th 8am
can anybody give me some info about the placements and prospects for MS(from bits)??

no i mean..like (i wanna get into the iits coz the us universities like standford, caltech. mostly select students from iits and nits.. for MS and stuff)...and how is BIT, Pilani in this regard ?

Is it? I didn't know IIT & NIT students are preferred over other students! I mean those foreign university guys should know BITS Pilani is much better than NITs and that the NITs, although good, are no match to the IITs or BITS Pilani (Rajasthan).

For transfer students, I believe IIT-M&D are most reputed.
For M.Sc, firstly, you should broaden your list. Those universities you listed are a very very very long shot. Its mostly what you DO at the universities at undergrad level that matters, rather than whether you went to IIT-Kanpur or Kharagpur.
Obviously, you don't want something like Babu Banarasi Das on your US applications...  :P

thts wat!
i ll try and keep with projects and stuff and stuff...
but i don want something like
iit-b(islamia institute of tech-bangalore) on my resume
(no offence to iit-b students)

Did any of you guys attempt the extra 12 questions or was there not enough time or was it difficult to attempt all the questions for fear of getting negative .

I still have my exam so please help me...  :oops:
And what is the cutoff for mech at pilani

got 366 cudn't attempt those fuking extra qs  :mad:

i had taken some risk to get to those, and in the end only 1 min was left sigh

neways can i get cs at pilani ???

atleast eee ???

@sumanth
on the contrary i felt english and logical reasoning to be easiest

[quote="Sagar_Vaze"]
@sumanth
on the contrary i felt english and logical reasoning to be easiest[/quote]

Even in one minute?

EDIT: waitaminute! isnt the english and logic q part of extra q or are they in core too?

last yr cut off for cs was 349 and mech was 324
so i guess every ones get wat they wants
and cs cut offs for last 3yrs :2006-368
2007-357
2008-349
so i reckon chetan might just make it..and sagar ofcourse u ll get cs! easy!

@sumanth
whr did u get the 2008 data ?
@conjurer
some questions were like
which is the misspelt word
auxillary [b]aquaintance[/b] sth else
besides there were three arrange the sentences to make coherent paragraph type question---a piece of cake if u solve some cat questions  :D

@sagar
its been put on ankur mittal
or just google for bitsat 2009 cut off
u ll get 2-3 sites
electrical and electronics-339
electronics and instrumentation-334
mechanical -324
and chemical and civil some 306 and 301

good

it as good as over 4 me :o  i messed up my DPMT paper full tym :mad:

i wont get even in top 20 :(

actually both will be fine...

as u can see i m already dead... :(

Sagar, wait for the results. You can get something good at IIT. Do not trouble trouble until trouble troubles you!

Whatever that means...

[quote="Sagar_Vaze"]so shud i take cs@bits or s**t@iit  :maybe:[/quote]
For a moment I thought you were abusing someone!  :rotfl:

@sagar
i spoke to a few of my seniors at iit
since i have chances of getting mech at madras
and i like mech..they said go ahead....
and they said...if u are not desperate for ny branch u shud go ahead with iit...
and they also said dropping an year is not worth it
so i donno...if i wer u ..i wud take a decent course in the top 5iits
(assuming tht u ll get a 3digit rank)


[color=darkblue]Fie un triunghi isoscel $ABC$ in care $AB=AC$ si punctele $O$,  $G$ sunt centrul cercului circumscris respectiv centrul de greutate al triunghiului. Fie mijlocul $D$ al laturii $[AB]$ si centrul de greutate $M$ al triunghiului $ACD$. Sa se arate ca $OD\perp MG$ si $MO\perp DG$ ([b]Autor[/b] : [u]Marcel Chirita[/u]).[/color]

[b][color=red][u]Este o problemă frumoasă !![/u][/color][/b]

[img]http://img.photobucket.com/albums/v218/hudrea/clasavii.jpg[/img]

[u][b]Soluţie:[/b][/u]

[hide="Indicaţie:"]$MG\left\| {AB} \right.$ şi $AG \bot DM$
[/hide]


Expreses as a single number in lowest terms:

$\frac{28 \cdot 26 \cdot 24 \cdot 22}{11 \cdot 12 \cdot 13 \cdot 14}$


algebra polynomial function calculus derivative factorial limit

Find all polynomials like $P(x)$ so that for every natural number like $n, P(n)=n!$.

We can't do an infinite taylor approximation either, because $x!$ has no derivative...
Or does it?  The Gamma Function?

maybe you can derive stirlings approximation of n!
but ZetaX's observation is correct and settles the problem

[quote="ZetaX"]Since $n!$ grows faster than any polynomial: there are none.[/quote]
could you prove this exactly?

You can make a similar argument for polynomial functions and exponential functions, though...but there still exists an infinite polynomial for the exponential.
But I suppose the gamma function is different...
If you approximate the derivative using Euler's method in reverse, you get that the derivative is very close to the factorial itself.

It can easily be proved that for every natural number like $k$, we have $\lim_{n \to \infty}\frac{n^{k}}{n!}=0$. Now suppose that a polynomial like $P$ exists so that $\deg P=k$ and for every natural number like $n, P(n)=n!$. Also suppose that $P(x)=\sum_{i=0}^{k}{a_{i}}{x^{i}}$. Then we would have $\lim_{n \to \infty}\frac{P(n)}{n^{k}}=a_{k}$, and $\lim_{n \to \infty}\frac{P(n)}{n!}=1$. This is a contradiction since $0.{a_{k}}\neq 1$(Because $\lim_{n \to \infty}\frac{n^{k}}{n!}. \lim_{n \to \infty}\frac{P(n)}{n^{k}}$ should equal $\lim_{n \to \infty}\frac{P(n)}{n!}$).