inequalities
calculus
inequalities proposed

I am happy to see you again at the forum, after a long time, with your very nice inequalities.

Thank you very much, my dear friend. :)

It's not easy. I have to solve it by my technique which called MIS.
The last ineq I obtain is:
$ 4abc\sum\ a^3 \plus{} 3\sum\ ab(a^4 \plus{} b^4) \plus{} \sum\ a^4(b^2 \plus{} c^2) \geq\ 2abc \sum\ ab(a \plus{} b) \plus{} 6a^2b^2c^2 \plus{} 6\sum\ a^3b^3$
Obviously trues by Am-Gm because:
$ 4abc \sum\ a^3 \geq\ 2abc \sum\ ab(a \plus{} b)$
$ 3 \sum\ a^5(b \plus{} c) \geq\ 6 \sum\ a^3b^3$
and  $ \sum\ a^4(b^2 \plus{} c^2) \geq\ 6a^2b^2c^2$
I'll post my full proof later. Wait for another proofs for it  :)

[quote="ductrung"]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac {a}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \frac {b}{(c \plus{} a)(b^2 \plus{} ca)} \plus{} \frac {c}{(a \plus{} b)(c^2 \plus{} ab)}\le\frac {9}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

By AM-GM easily to see that
\[ \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}} \le \frac {9}{{4\left( {ab \plus{} bc \plus{} ca} \right)}}
\]
Hence it suffices to prove the stronger
\[ \sum\limits_{cyc} {\frac {a}{{\left( {b \plus{} c} \right)\left( {a^2 \plus{} bc} \right)}}} \le \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}}
\]

\[ \sum\limits_{cyc} {\frac {1}{{\left( {b \plus{} c} \right)^2 }}} \minus{} \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}} \le \sum\limits_{cyc} {\frac {1}{{\left( {b \plus{} c} \right)^2 }}} \minus{} \sum\limits_{cyc} {\frac {a}{{\left( {b \plus{} c} \right)\left( {a^2 \plus{} bc} \right)}}}
\]

\[ \sum\limits_{cyc} {\frac {{\left( {b \minus{} c} \right)^2 }}{{\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}} \le 2\sum\limits_{cyc} {\frac {{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}}{{\left( {b \plus{} c} \right)^2 \left( {a^2 \plus{} bc} \right)}}}
\]

\[ \sum\limits_{cyc} {X\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]
where
\[ X \equal{} \frac {1}{{\left( {b \plus{} c} \right)^2 }}\left[ {\frac {2}{{a^2 \plus{} bc}} \minus{} \frac {1}{{\left( {a \plus{} b} \right)^2 }} \minus{} \frac {1}{{\left( {a \plus{} c} \right)^2 }}} \right]
\]
or we can rewrite it to
\[ X \equal{} \frac {{2\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \minus{} \left( {a \plus{} b} \right)^2 \minus{} \left( {a \plus{} c} \right)^2
\]
Note that we have the follow identity
\[ \sum\limits_{cyc} {\left( {b \minus{} c} \right)^2 \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \equal{} 0
\]
By the identity and note that
\[ \frac {{2\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \minus{} \left( {a \plus{} b} \right)^2 \minus{} \left( {a \plus{} c} \right)^2 \equal{} \frac {{2a^2 \left( {b \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \plus{} 2a\left( {b \plus{} c} \right) \minus{} \left( {b \minus{} c} \right)^2
\]
Hence we can rewrite the inequality as
\[ \sum\limits_{cyc} {\left[ {\frac {{2a^2 \left( {b \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \plus{} 2a\left( {b \plus{} c} \right)} \right]} \left( {a \minus{} b} \right)\left( {a \minus{} c} \right) \ge 0
\]
\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \cdot \sum\limits_{cyc} {\frac{a}{{a^2  \plus{} bc}} \cdot \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}  \ge 0
\]
\[ \sum\limits_{cyc} {\frac{a}{{a^2  \plus{} bc}} \cdot \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}  \ge 0
\]
Without loss of generality we may assume $ a \ge b \ge c$.
Then easily to prove that
\[ a \cdot \frac{a}{{a^2  \plus{} bc}} \ge b \cdot \frac{b}{{b^2  \plus{} ca}}
\]
Finally by Vornicu Schur the inequality is proved. :)
For the stronger we have equality when $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$. :)

[quote="ductrung"]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac {a}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \frac {b}{(c \plus{} a)(b^2 \plus{} ca)} \plus{} \frac {c}{(a \plus{} b)(c^2 \plus{} ab)}\le\frac {9}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

Welcom you go back, ductrung :)

Your inequality is nice, but it's wecker than the following one of mine

$ \frac{a(b\plus{}c)}{a^2\plus{}bc}\plus{}\frac{b(c\plus{}a)}{b^2\plus{}ca}\plus{}\frac{c(a\plus{}b)}{c^2\plus{}ab}\le1\plus{}\frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{4abc}$
:)

[quote="dduclam"]
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$
:)[/quote]
If $ a,b,c$ be the sidelenghts of a triangle then
\[ 1 \plus{} 2 \cdot \frac{{ab \plus{} bc \plus{} ca}}{{a^2  \plus{} b^2  \plus{} c^2 }} \le \frac{{a\left( {b \plus{} c} \right)}}{{a^2  \plus{} bc}} \plus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \plus{} \frac{{c\left( {a \plus{} b} \right)}}{{c^2  \plus{} ab}} \le 3
\]
:)

[quote="Honey_S"][quote="ductrung"]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac {a}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \frac {b}{(c \plus{} a)(b^2 \plus{} ca)} \plus{} \frac {c}{(a \plus{} b)(c^2 \plus{} ab)}\le\frac {9}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

By AM-GM easily to see that
\[ \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}} \le \frac {9}{{4\left( {ab \plus{} bc \plus{} ca} \right)}}
\]
Hence it suffices to prove the stronger
\[ \sum\limits_{cyc} {\frac {a}{{\left( {b \plus{} c} \right)\left( {a^2 \plus{} bc} \right)}}} \le \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}}
\]

\[ \sum\limits_{cyc} {\frac {1}{{\left( {b \plus{} c} \right)^2 }}} \minus{} \sum\limits_{cyc} {\frac {1}{{\left( {a \plus{} b} \right)\left( {a \plus{} c} \right)}}} \le \sum\limits_{cyc} {\frac {1}{{\left( {b \plus{} c} \right)^2 }}} \minus{} \sum\limits_{cyc} {\frac {a}{{\left( {b \plus{} c} \right)\left( {a^2 \plus{} bc} \right)}}}
\]

\[ \sum\limits_{cyc} {\frac {{\left( {b \minus{} c} \right)^2 }}{{\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}} \le 2\sum\limits_{cyc} {\frac {{\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)}}{{\left( {b \plus{} c} \right)^2 \left( {a^2 \plus{} bc} \right)}}}
\]

\[ \sum\limits_{cyc} {X\left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]
where
\[ X \equal{} \frac {1}{{\left( {b \plus{} c} \right)^2 }}\left[ {\frac {2}{{a^2 \plus{} bc}} \minus{} \frac {1}{{\left( {a \plus{} b} \right)^2 }} \minus{} \frac {1}{{\left( {a \plus{} c} \right)^2 }}} \right]
\]
or we can rewrite it to
\[ X \equal{} \frac {{2\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \minus{} \left( {a \plus{} b} \right)^2 \minus{} \left( {a \plus{} c} \right)^2
\]
Note that we have the follow identity
\[ \sum\limits_{cyc} {\left( {b \minus{} c} \right)^2 \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \equal{} 0
\]
By the identity and note that
\[ \frac {{2\left( {a \plus{} b} \right)^2 \left( {a \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \minus{} \left( {a \plus{} b} \right)^2 \minus{} \left( {a \plus{} c} \right)^2 \equal{} \frac {{2a^2 \left( {b \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \plus{} 2a\left( {b \plus{} c} \right) \minus{} \left( {b \minus{} c} \right)^2
\]
Hence we can rewrite the inequality as
\[ \sum\limits_{cyc} {\left[ {\frac {{2a^2 \left( {b \plus{} c} \right)^2 }}{{a^2 \plus{} bc}} \plus{} 2a\left( {b \plus{} c} \right)} \right]} \left( {a \minus{} b} \right)\left( {a \minus{} c} \right) \ge 0
\]

\[ \left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right) \cdot \sum\limits_{cyc} {\frac {a}{{a^2 \plus{} bc}} \cdot \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]

\[ \sum\limits_{cyc} {\frac {a}{{a^2 \plus{} bc}} \cdot \left( {a \minus{} b} \right)\left( {a \minus{} c} \right)} \ge 0
\]
Without loss of generality we may assume $ a \ge b \ge c$.
Then easily to prove that
\[ a \cdot \frac {a}{{a^2 \plus{} bc}} \ge b \cdot \frac {b}{{b^2 \plus{} ca}}
\]
Finally by Vornicu Schur the inequality is proved. :)
For the stronger we have equality when $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$. :)[/quote]

Very nice your proof, Honey_S, congratulations and thank you very much. :)

[quote="Dimitris X"][quote="nguoivn"] I have to solve it by my technique which called MIS.
[/quote]
Can you explain?????[/quote]
You can see the main hint of this method in my following proof:
But with this method, I have to calculus lot. So, I rarely want to use it.  :oops:
PS: I like Honey_S's proof. And I really want to see your proof for this ineq: http://www.mathlinks.ro/Forum/viewtopic.php?t=255583
I have to used MIS again  :(

Nice proof, Honey_S  :clap2: I proved the stronger inequality using SOS but the proof is not as nice as yours. Basically, we have
$ \frac{2}{(a\plus{}b)(a\plus{}c)}\minus{}\frac{b}{(c\plus{}a)(b^2\plus{}ca)}\minus{}\frac{c}{(a\plus{}b)(c^2\plus{}ab)}$
$ \equal{} \frac{a(b\plus{}c\minus{}a)(b\minus{}c)^2}{(a\plus{}b)(a\plus{}c)(b^2\plus{}ca)(c^2\plus{}ab)}$
and so it suffices to prove that
$ A(b\minus{}c)^2\plus{}B(c\minus{}a)^2\plus{}C(a\minus{}b)^2 \ge 0$
where
$ A\equal{}a(b\plus{}c)(a^2\plus{}bc)(b\plus{}c\minus{}a)$

[quote="dduclam"]
Welcom you go back, ductrung :)

Your inequality is nice, but it's wecker than the following one of mine

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$
:)[/quote]

Thank you, Lam. Why this inequality is stronger? Anyway, it is weaker than
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}$
which can be found at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/url] post [b]#12[/b]; not to mention the left-hand side of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135642[/url]. And how about this one [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135645[/url]: this is to you, Lam  :wink:

[quote="ductrung"]
[quote="dduclam"]
Welcom you go back, ductrung :)

Your inequality is nice, but it's wecker than the following one of mine

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$
:)[/quote]

Thank you, Lam. Why this inequality is stronger?
[/quote]
Yes, I also don't understand why it is stronger. Please more detail  :oops:
And as Duc said, it's easy because:
LHS $ \leq\ \frac{(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca}$  (by Voni Schur)
and $ \frac{(a\plus{}b\plus{}c)^2}{ab\plus{}bc\plus{}ca} \leq\ 1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$  (by SOS)

[quote="ductrung"]
And how about this one [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135645[/url]: this is to you, Lam  :wink:[/quote]
It's a nice application of Am-Gm, my friend. I think your proof also used Am-Gm:
$ \frac{ab(a\plus{}c)(b\plus{}c)}{(a^2\plus{}bc)(b^2\plus{}ca)} \leq\ 1$
And $ (a^2\plus{}bc)^2 \geq\ 4a^2bc$
 :)

For $ a,b,c$ positive reals

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$

proof:

$ LHS \equal{}\sum \frac {a(b \plus{} c)}{a^2 \plus{} bc} \leq \sum \frac {a(b \plus{} c)\sqrt{bc}}{2abc} \leq
\sum \frac {a(b \plus{} c)^2}{4abc} \equal{} \sum \frac {a^2(b \plus{} c)\plus{}2abc}{4abc} \equal{} RHS$

[quote="ductrung"]
Thank you, Lam. Why this inequality is stronger? Anyway, it is weaker than
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le\frac {(a \plus{} b \plus{} c)^2}{ab \plus{} bc \plus{} ca}$
which can be found at [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=116853[/url] post [b]#12[/b]; not to mention the left-hand side of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135642[/url]. [/quote]

Dear Ductrung and nhocnhoc,
Sorry for my delay. I'm too busy. :blush:

The first, I will explain why your Inequality is weaker than my one, Ductrung.
Indeed, your Inequality is equivalent to
$ \sum\frac {4abc}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \sum\frac {4a^2}{a^2 \plus{} bc}\le9$
$ \Leftrightarrow \sum\frac {4abc}{(b \plus{} c)(a^2 \plus{} bc)}\le1 \plus{} \frac {4abc(a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)}{(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)}$  (1)
And my Inequality is equivalent to
$ \sum\frac {(a \plus{} b)(a \plus{} c)}{a^2 \plus{} bc}\le4 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$
$ \Leftrightarrow \sum\frac {4abc}{(b \plus{} c)(a^2 \plus{} bc)}\le1 \plus{} \frac {16abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$       (2)
Using my other Inequality : $ (a^3 \plus{} b^3 \plus{} c^3 \plus{} abc)(a \plus{} b)(b \plus{} c)(c \plus{} a)\ge4(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab)$
(here http://www.mathlinks.ro/viewtopic.php?t=244158 ) then (1) weaker than (2). :)

[quote="ductrung"]And how about this one [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135645[/url]: this is to you, Lam  :wink:[/quote]

Thank for your nice problem, see my solution (as nhocnhoc'method): http://www.mathlinks.ro/Forum/viewtopic.php?p=1399069#1399069

[quote="manlio"]For $ a,b,c$ positive reals

$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le1 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc}$

proof:

$ LHS \equal{} \sum \frac {a(b \plus{} c)}{a^2 \plus{} bc} \leq \sum \frac {a(b \plus{} c)\sqrt {bc}}{2abc} \leq \sum \frac {a(b \plus{} c)^2}{4abc} \equal{} \sum \frac {a^2(b \plus{} c) \plus{} 2abc}{4abc} \equal{} RHS$[/quote]

Nice, manlio :) . Infac, we have stronger Inequality as follows
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le2 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}$
:)

@ddlam: Thank you, my brother for an interesting answers. Now, I see  :)
[quote="dduclam"]Similar Duc'Inequality, the following Inequality is also true and nice

For all nonnegative real numbers $ a,b$  and $ c$, no two of which are zero, we have
\[ \frac {a}{(b \plus{} c)(4a^2 \plus{} 5bc)} \plus{} \frac {b}{(c \plus{} a)(4b^2 \plus{} 5ca)} \plus{} \frac {c}{(a \plus{} b)(4c^2 \plus{} 5ab)}\le\frac1{2(ab \plus{} bc \plus{} ca)}
\]
Equality hold for $ a \equal{} b \equal{} c$ or $ a \equal{} b,c \equal{} 0$ or any cyclic permutation. :)[/quote]
Equality hold for $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$. My method above still solved it, of course, not nice  :oops:

Yes, my younger brother. This is my typing mistake.  :blush:

@nhocnhoc: My nick name is dduclam, not ddlam  :lol:

Dear, Lam. Thank you for your reply.

[quote="dduclam"] Infac, we have stronger Inequality as follows
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le2 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}$
:)[/quote]

But this is [b]still weaker[/b] than [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135645[/url], agree?

[quote="dduclam"] We have stronger Inequality as follows
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le2 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}$
:)[/quote]

Can you please tell me how to solve this nice inequality in a simple way?

I tried by Vornicu-Shur but I failed :(

Thank you very much, dduclam. :)

[quote="dduclam"]
Infact, we have stronger Inequality as follows
$ \frac {a(b \plus{} c)}{a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} ab}\le2 \plus{} \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{8abc}$
:)[/quote]

[quote="manlio"]
Can you please tell me how to solve this nice inequality in a simple way?
[/quote]

You can prove by this way, very simple. :)

Assume $ a \ge b \ge c$ then
\[ 2 \minus{} \frac{{a\left( {b \plus{} c} \right)}}{{a^2  \plus{} bc}} \minus{} \frac{{c\left( {a \plus{} b} \right)}}{{c^2  \plus{} ab}} \ge 2 \minus{} \frac{{a\left( {b \plus{} c} \right)}}{{a\left( {b \plus{} c} \right)}} \minus{} \frac{{c\left( {a \plus{} b} \right)}}{{c\left( {a \plus{} b} \right)}} \equal{} 2 \minus{} 2 \equal{} 0
\]
Hence
\[ \frac{{a\left( {b \plus{} c} \right)}}{{a^2  \plus{} bc}} \plus{} \frac{{c\left( {a \plus{} b} \right)}}{{c^2  \plus{} ab}} \le 2
\]
Moreover
\[ \frac{{\left( {a \plus{} b} \right)\left( {b \plus{} c} \right)\left( {c \plus{} a} \right)}}{{8abc}} \minus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \equal{} \frac{{\left( {c \plus{} a} \right)\left( {b^2  \plus{} ca \plus{} b\left( {c \plus{} a} \right)} \right)}}{{8abc}} \minus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \ge \frac{{\left( {c \plus{} a} \right)\left( {b^2  \plus{} ca} \right)}}{{4abc}} \minus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \equal{} \frac{{\left( {c \plus{} a} \right)\left( {b^2  \plus{} ca} \right)^2 }}{{4abc\left( {b^2  \plus{} ca} \right)}} \minus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \ge \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \minus{} \frac{{b\left( {c \plus{} a} \right)}}{{b^2  \plus{} ca}} \equal{} 0
\]
This completes the proof. :)

Thank you very much, Honey_S, for your nice proof. :)

[quote="ductrung"]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac {a}{(b \plus{} c)(a^2 \plus{} bc)} \plus{} \frac {b}{(c \plus{} a)(b^2 \plus{} ca)} \plus{} \frac {c}{(a \plus{} b)(c^2 \plus{} ab)}\le\frac {9}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

Dear ductrung, can you please post your solution for your nice inequality.

Thank you very much. :)

[quote="dduclam"]Similar Duc'Inequality, the following Inequality is also true and nice

For all nonnegative real numbers $ a,b$  and $ c$, no two of which are zero, we have
\[ \frac {a}{(b \plus{} c)(4a^2 \plus{} 5bc)} \plus{} \frac {b}{(c \plus{} a)(4b^2 \plus{} 5ca)} \plus{} \frac {c}{(a \plus{} b)(4c^2 \plus{} 5ab)}\le\frac1{2(ab \plus{} bc \plus{} ca)}
\]
Equality hold for $ a \equal{} b \equal{} c$ or $ abc \equal{} 0$ . :)[/quote]

This inequality, I have proposed and solved it nearly one year ago. I created when trying to find a proof by Cauchy Schwarz for the following nice inequality of Vasc :)
\[ \sqrt{\frac{x}{4x\plus{}5y}}\plus{}\sqrt{\frac{y}{4y\plus{}5z}}\plus{}\sqrt{\frac{z}{4z\plus{}5x}} \le 1\]
for $ x,y,z>0.$

:)

[quote="manlio"]Dear ductrung, can you please post your solution for your nice inequality. [/quote]
Dear Manlio, my first proof uses SOS and thus is not nice. Below is the latest: As Honey_S noted, it is a well known fact that
$ \frac {2(a \plus{} b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\le\frac {9}{4(ab \plus{} bc \plus{} ca)}$
so it suffices to prove
$ \sum_{cyc}\frac {a}{(b \plus{} c)(a^2 \plus{} bc)}\le\frac {2(a \plus{} b \plus{} c)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
Multiplying thru by $ (a \plus{} b)(b \plus{} c)(c \plus{} a)$ we obtain
$ \sum_{cyc}\frac {a(a \plus{} b)(a \plus{} c)}{a^2 \plus{} bc}\le 2(a \plus{} b \plus{} c)$
$ \sum_{cyc}\frac {a[a^2 \plus{} bc \plus{} a(b \plus{} c)]}{a^2 \plus{} bc} \le 2(a \plus{} b \plus{} c)$
$ \sum_{cyc}\left( a \plus{} \frac {a^2(b \plus{} c)}{a^2 \plus{} bc} \right) \le 2(a \plus{} b \plus{} c)$
$ \sum_{cyc}\frac {a^2(b \plus{} c)}{a^2 \plus{} bc}\le a \plus{} b \plus{} c$
Now we have a well-known inequality: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=159311[/url] post [b]#2[/b] (i posted the solution to this general inequality in a .pdf file somewhere but couldn't find it).

[quote=ductrung]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac{a}{(b\plus{}c)(a^2\plus{}bc)}\plus{}\frac{b}{(c\plus{}a)(b^2\plus{}ca)}\plus{}\frac{c}{(a\plus{}b)(c^2\plus{}ab)}\le\frac{9}{4(ab\plus{}bc\plus{}ca)} \][/quote]
The following inequality is also true
[img]http://s8.sinaimg.cn/middle/0018zOAxgy7247z2hSfa7&690[/img]

[quote=luofangxiang]
The following inequality is also true
[img]http://s8.sinaimg.cn/middle/0018zOAxgy7247z2hSfa7&690[/img][/quote]

We have
\[2\sum{a}-\prod{(a+b)}\sum{\frac{a}{(b+c)(a^2+bc)}}=\frac{abc\sum{(a^2-b^2)^2}}{2\prod{(a^2+bc)}}\ge{0}\]

Try Cauchy Schwarz-Muirhead-Chebysev.

Some awesome things happens.

Thanks for this nice inequality.

We can rewrite the last inequality in the following form $\sum_{cyc}\frac{a^2(b+c)}{a^2+bc}\leq a+b+c$ and to use a full expanding.

Let $a$, $b$ and $c$ be positive numbers and $k\geq2$. Prove that:
$$\frac{a^k(b+c)}{a^2+bc}+\frac{b^k(a+c)}{b^2+ac}+\frac{c^k(a+b)}{c^2+ab}\leq a^{k-1}+b^{k-1}+c^{k-1}$$

[quote=luofangxiang]Don't need Cauchy Schwarz-Muirhead-Chebysev；Directly can prove[/quote]

Yes,but it was really nice.
Not so hard to try :pilot:

[quote=ductrung]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac{a}{(b\plus{}c)(a^2\plus{}bc)}\plus{}\frac{b}{(c\plus{}a)(b^2\plus{}ca)}\plus{}\frac{c}{(a\plus{}b)(c^2\plus{}ab)}\le\frac{9}{4(ab\plus{}bc\plus{}ca)} \][/quote]

only three lines!!!

[quote=ductrung]Prove that for all nonnegative real numbers $ a$, $ b$ and $ c$, no two of which are zero,
\[ \frac{a}{(b\plus{}c)(a^2\plus{}bc)}\plus{}\frac{b}{(c\plus{}a)(b^2\plus{}ca)}\plus{}\frac{c}{(a\plus{}b)(c^2\plus{}ab)}\le\frac{9}{4(ab\plus{}bc\plus{}ca)} \][/quote]

Multiplying the both sides by $(a+b)(b+c)(c+a)$, so we must prove

$$\frac{a(a+b)(a+c)}{a^2+bc}+\frac{b(b+a)(b+c)}{b^2+ca}+\frac{c(c+a)(c+a)}{c^2+ab} \le \frac{9(a+b)(b+c)(c+a)}{4(ab+bc+ca)}$$

It suffices to prove that

$$\frac{a(a+b)(a+c)}{a^2+bc}+\frac{b(b+a)(b+c)}{b^2+ca}+\frac{c(c+a)(c+a)}{c^2+ab} \le 2(a+b+c) \le \frac{9(a+b)(b+c)(c+a)}{4(ab+bc+ca)}$$

The right side inequality is true by AM-GM and well known $(a+b)(b+c)(c+a) \ge \frac{8}{9}(a+b+c)(ab+bc+ca)$

The left side inequality is true also because

$$2(a+b+c)- \left(       \frac{a(a+b)(a+c)}{a^2+bc}+\frac{b(b+a)(b+c)}{b^2+ca}+\frac{c(c+a)(c+a)}{c^2+ab}    \right) =\frac{abc\left(a^2-b^2\right)^2+abc\left(b^2-c^2\right)^2+abc\left(c^2-a^2\right)^2}{2\left(a^2+bc\right)\left(b^2+ca\right)\left(c^2+ab\right)} \ge 0$$

Done.


algebra
function
domain
calculus
derivative
modular arithmetic
real analysis

[quote="harazi"]Where is the function $ f(x)=\sum_{n\geq 1} {\frac{sin(n^2x)}{n^2}}$ derivable?[/quote]

 What do you mean by " derivable "? Is it differentiable ( having derivative ) or something else? :?

I'll assume that this series of functions is summed to infinity.

First of all, to take the derivative of a series we need to know if the function: $\sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2}$  converges pointwise and that it's derivative $\sum_{n=1}^{\infty} \cos(n^2 x)$ converges uniformly.

To do this we use the Weierstrass M-test:

$\frac{\sin(n^2 x)}{n^2} \leq \frac{1}{n^2} = M_{n}$

Now we check the convergence of $M_{n}$.

$\sum_{n=1}^{\infty} M_{n} = \sum_{n=1}^{\infty} \frac{1}{n^2}$

This obviously converges (uniformly and thus pointwise). But the series $\sum_{n=1}^{\infty} \cos(n^2 x)$ does not converge uniformly. So you cannot take the derivative as follows:

$\left(\sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2}\right)' = \sum_{n=1}^{\infty} \left(\frac{\sin(n^2 x)}{n^2}\right)'$

Excuse me, but you proof, swag, is one big mistake.

[quote="swag"]First of all, to take the derivative of a series we need to know if the function: $\sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2}$  is uniformly convergent.[/quote]
It is not a sufficient condition.

[quote]This obviously converges. So the original series of functions converges uniformly and so we can take it's derivative as follows:
$\left( \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2} \right)' = \left( \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2}' \right) =  \left( \sum_{n=1}^{\infty} 2\,{\frac {\cos \left( {n}^{2}x \right) x}{n}}-2\,{\frac {\sin \left( {n}^{2}x \right) }{{n}^{3}}} \right)$[/quote]
Your formula is totally wrong. Right way to differentiate series (though it is wrong equality):
\[\left( \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2} \right)'=\sum_{n\geq 1}\cos(n^2x).\]

Regards,
Myth

[quote="Myth"]
It is not a sufficient condition.[/quote]

You're right, I fixed that.

[quote]
Your formula is totally wrong. Right way to differentiate series (though it is wrong equality):
\[\left( \sum_{n=1}^{\infty} \frac{\sin(n^2 x)}{n^2} \right)'=\sum_{n\geq 1}\cos(n^2x).\][/quote]

My mistake, I don't know why I was taking the derivative with respect to n. I fixed that bracket too.

I don't have any progress to report, but I will make a conjecture. I conjecture that the function is differentiable everywhere except zero. Furthermore, the function $\sum\frac{z^{n^2}}{n^2}$ can be analytically continued to a larger domain than the unit disc.

All right, I'm guessing.

The conjecture is false. The function is derivable except a numerable set. This is what makes it hyper interesting.

[quote="harazi"]The function is derivable except a numerable set.[/quote]
To be technical, when I meant "everywhere exept zero" I really meant "for all $x\not\equiv0\pmod{2\pi}.$" In the quote above, do you mean that the set on which the function is not differentiable is (countably) infinite in the set $[0,2\pi)$?

Yes, the points in which f is non-derivable are quite strange. After 4 days, when I return France I will tell you exactly in which points f is non differentiable. I found a material there.

Sorry for the very late answer, it is just now that I found my material. It is written there that this damn function has a derivative which equals $-\frac{1}{2}$ in any point of the form $\pi\cdot\frac{2p+1}{2q+1}$ with p,q integers and only in these points. However, they do not give the proof , saying that a proof was found by Hardy and Gerver.


LaTeX logarithms limit calculus function inequalities induction

Hi guys,
I'm a 14-year-old Italian boy, attending the first year of high school...

Some time ago, thinking that 4^2 = 2^4,
I tried to find other couples of numbers with that property (x^y = y^x).

Since I was not able to find any solution for x^y = y^x,
I assumed y = x^k , where k is a constant

So, the equation becomes:

x^(x^k) = x^(xk)

x^k = xk

x^(k-1) = k

and the solution is

x = k^(1/(k-1)),

and

y = x^k = (k^k)^(1/(k-1)) = k^(k/(k-1))

This is useful to find the couple of numbers, picking a random k.

... I'm not totally sure that this solution is correct...
In facts, k mustn't be 1, that means x not-equal y, even if it is a real solution...

But this isn't my only doubt: are all the couples of numbers made in this way? And is it the only solution? Except x=y, are only 2 and 4 (and vice-versa) the integers which satisfy the equation (I guess yes, because the only way to get y integer while x is integer is k/(k-1) be integer, which leads k=2)...

I asked my maths teacher, but he didn't help me a lot... I also searched on the net, but the solution is too difficult for me...

I'm sorry for my English, I promise I'll improve it, and I'll learn LaTex, too.

Thank you in advance,

Gabriele  :)

Yes, all of the solutions are of this form, at least if you restrict $ x, y > 0$.  Are you familiar with logarithms?  There is in fact always a constant $ k$ such that $ y \equal{} x^k$.

[quote="Gabuntu94"]Except x=y, are only 2 and 4 (and vice-versa) the integers which satisfy the equation (I guess yes, because the only way to get y integer while x is integer is k/(k-1) be integer, which leads k=2)... [/quote]
That is the only integers satisfying the equation. The proof is a bit more complicated though. E.g. $ 25^{1/2}$ is also an integer, so there could be an integer solution with $ \frac{k}{k\minus{}1} \notin \mathbb{Z}$.

[quote="Gabuntu94"]... I'm not totally sure that this solution is correct...
In facts, k mustn't be 1, that means x not-equal y, even if it is a real solution...

But this isn't my only doubt: are all the couples of numbers made in this way? And is it the only solution? Except x=y, are only 2 and 4 (and vice-versa) the integers which satisfy the equation (I guess yes, because the only way to get y integer while x is integer is k/(k-1) be integer, which leads k=2)...[/quote]
[hide="Fixing the k=1 dilemma"]$ x^{k\minus{}1}\equal{}k$ implies that $ x \equal{} k^{\dfrac{1}{k\minus{}1}}$ only if $ k \neq 1$.

If $ k\equal{}1$, ie $ x\equal{}y$, then the equation becomes $ x^0 \equal{} k$, and thus $ k\equal{}1$.

Raising both sides to the $ \dfrac{1}{k\minus{}1}$th power is only valid when $ k \neq 1$. Otherwise, it is like dividing by zero. [/hide]

[hide="Proof that the numbers come in pairs"]$ x^y \equal{} y^x$ AND $ x \neq y$. (WLOG, assume that $ x<y$, since we can always switch $ x$ and $ y$ afterwards.)

$ y\ln(x) \equal{} x\ln(y)$

$ \dfrac{ln(x)}{x} \equal{} \dfrac{ln(y)}{y}$

Let $ f(x) \equal{} \dfrac{ln(x)}{x}$. We are looking for real positive pairs $ (x,y)$ such that $ f(x) \equal{} f(y)$.

Note that $ f(x)$ is continuous and differentiable over $ (1,\infty)$.

Also, $ f'(x) \equal{} \dfrac{1 \minus{} ln(x)}{x^2} \equal{} 0$ for only $ ln(x) \equal{} 1 \Rightarrow x \equal{} e$.

And, $ f'(x) > 0$ for all $ x$ in the interval $ (1,e)$, and $ f'(x) < 0$ for all $ x$ in the interval $ (e,\infty)$.

Furthermore, $ f(1) \equal{} 0$, and $ \lim_{x \to \infty}f(x) \equal{} 0$.

Thus, on the interval $ (1,\infty)$ the graph of $ f(x)$ looks like a "hill" that starts at $ y\equal{}0$ rises to $ f(e) \equal{} \dfrac{1}{e}$, and then slowly approaches $ y\equal{}0$.

Draw a horizontal line between $ y\equal{}0$, and $ y\equal{}f(e) \equal{} \dfrac{1}{e}$. This line intersects the graph of $ f(x)$ at two points. One before the peak at $ x\equal{}e$, and one after the peak.

So, if we assume that $ x<y$, then $ x$ will be a number in the interval $ (1,e)$ and $ y$ will be a number on the interval $ (e,\infty)$.

Thus, if $ x \neq y$ then there is always exactly one solution $ y>x$ except for $ x\equal{}e$, in which case $ y\equal{}e$.

Note: If you didn't understand the calculus part, just graph the function $ f(x) \equal{} \dfrac{\ln(x)}{x}$ over the interval $ (1,\infty)$ and observe the key points.
[/hide]

[hide="Proof that (x,y)=(2,4) is the only integer solution"]As we proved earlier, if $ x<y$, then $ x$ is in the interval $ (1,e)$. Since $ e \approx 2.71828$, the only integer in that interval is $ x\equal{}2$, and its corresponding solution is $ y\equal{}4$. Therefore, $ (x,y) \equal{} (2,4)$ is the only integer solution.[/hide]

consider the pair $ (x,y)$
let $ x\equal{}ky$ for some k,
 substituting it, we get
$ (ky)^{y}\equal{}(y)^{ky}$
$ ky\equal{}y^k$
$ y^{k\minus{}1}\equal{}k$
$ y\equal{}k^{\frac{1}{k\minus{}1}}$
$ x\equal{}k.k^{\frac{1}{k\minus{}1}}$
$ x\equal{}k^{\frac{k}{k\minus{}1}}$

I'll try to help you, Gabuntu94. When I tried to solve that problem I was 15 and I couldn't solve it in integers. Than I read it's solution and everything was clear. I hope it will be clear for you too.  :)
   $ x^{y} \equal{} y^{x}$ is equivalent with $ x^{\frac {1}{x}} \equal{} y^{\frac {1}{y}}$. Then consider the function $ f(n) \equal{} n^{\frac {1}{n}}$, where $ n$ is a positive integer. Using Bernoulli inequality and induction you can prove that this function is strictly decreasing after $ n \equal{} 4$ or something like this (maybe from $ n \equal{} 5$, I don't remember). Induction will be as the following: Our claim is $ f(n \plus{} 1) < f(n)$, where $ n \geq 4$. After this we have tu check for $ x \equal{} 1,2,3,4$ and it's not hard.
   Bernoulli Inequality states that: $ (1 \plus{} x)^{n} \geq {1 \plus{} xn}$, where $ x > { \minus{} 1}$, $ n > 0$ is an integer.

Can we do it like this?

Let $ x>y$

Then obviously $ x=ky,k\in\mathbb{Z}$

$ k^yy^y=y^ky^y$

$ k^y=y^k$ So if $ (x,y)$ is a solution such that $ x>y$, then ${ (\frac{x}{y},y})$ is a solution.

We can reduce the numbers like this until $ x=y$, or one of the numbers becomes 1.

If the numbers become equal, step before that is $ (x^2,x)$ Now we easily get $ x^2x^x=x^xx^x$, or $ x=2$.

If one number becomes 1, step before that is that the numbers are equal, first case.

So the only solution is $ (2,4),(4,2)$, excluding possibility $ x=y$.

Is this correct?

[quote="Bugi"]Can we do it like this?

Let $ x > y$

Then obviously $ x \equal{} ky,k\in\mathbb{Z}$[/quote]
Because?

If $ x>y$, then if $ y$'s got some prime factor which doesn't divide $ x$, then it will divide $ x^y$, but that's a contradiction.

[quote="Bugi"]If $ x > y$, then if $ y$'s got some prime factor which doesn't divide $ x$, then it will divide $ x^y$, but that's a contradiction.[/quote]
Good.
$ y^x\equal{}x^y \iff y^{ky} \equal{} k^y y^y$. Remember that $ y^ky^y \equal{} y^{k\plus{}y}$, so you cannot conclude $ y^k y^y \equal{} k^y y^y$. What you can conclude is $ y^k \equal{} ky \iff y^{k\minus{}1}\equal{}k \iff y \equal{} \sqrt[k\minus{}1]{k}$ which also leads to a solution since $ y,k \in \mathbb{N}$

Ah, I see thank you.

I thought $ (x)^{ab}\equal{}x^a\cdot x^b$...

Thank you guys very much!
Thank you everyone!

My doubt is finally over!!

Thank you again,
Gabriele  :)


[quote="frt"][hide="hint"]
Using area ratio, we can immediately find $AP: PE$ and $AF: FC$.
Then, we'll get a pair of equations regarding the areas of $\triangle ADP$, $\triangle BDP$ and $\triangle BEP$.
Alternatively, we can directly find $BE: EC$. This way is probably faster.[/hide][/quote]

hmm...returning to aops?

anyway...mass points might be the way to go here...or ceva/menelaus..etc

or use 1/2bh formula for stuff to get length ratios

these problems are all the same

[quote="Quevvy"]Given triangles ABC with D on AB, E on BC and F on AC such that AE, BF and CD intersect at P.

If the area of triangle AFP is 40, the area of triangle CFP is 30 and the area of triangle CEP is 35, what is the area of triangle BPD?[/quote]

I used areas according to base theorem and SEVA theorem and found
 [hide]84[/hide]