(1) $ f(x) \equal{} x^3 \plus{} ax^2 \plus{} ax \plus{} 1$ has local extrema at $ x \equal{} \alpha ,\ \beta (\alpha <\beta)$. Find $ a,\ \alpha$ such that $ f(\alpha) \plus{} f(\beta) \equal{} 2$.

(2) Find $ b,\ c$ such that $ \int_1^x g(t)\ dt \equal{} x^2 \plus{} bx \plus{} c,\ g(1) \equal{} 5$.

hello
I hope this would be correct. :|
[hide="2"]
differentiating both the sides ,
$ g(x)\equal{}2x\plus{}b$
b=3
$ \int^{x}_{1}(2x\plus{}3)dx\equal{}(x^2\plus{}3x\plus{}C)^{x}_{1}\equal{}x^2\plus{}3x\minus{}4$
$ c\equal{}\minus{}4$
[/hide]
thank u

O.K. :lol:

Plugg $ x\equal{}1$ into the integral equation gives ...

hello
[hide="1"]
$ \alpha\plus{}\beta\equal{}\frac{\minus{}2a}{3}$,$ \alpha\beta\equal{}\frac{a}{3}$
 $ \alpha^3\plus{}\beta^3\plus{}a(\alpha^2\plus{}\beta^2\plus{}\alpha\plus{}\beta)\equal{}0$
 $ (\alpha\plus{}\beta)^3\minus{}3\alpha\beta(\alpha\plus{}\beta)\plus{}a((\alpha\plus{}\beta)^2\plus{}\alpha\plus{}\beta\minus{}2\alpha\beta)\equal{}0$
putting the value we will get
 $ \frac{4a^3}{27}\minus{}\frac{2a^2}{3}$=0
$ a\equal{}0,4.5$
[/hide]
thank u

What's the range of $ a$ such that $ f(x)$ has local extrema?


calculus
integration
function
real analysis
real analysis theorems

Anyone do have an idea of iterated double Stieljes Integral?
Let's say $g(u,v)=h_{1}(u)h_{2}(v)~~~~\forall u\in [a,b],~~\forall v\in [c,d]$, do we have

${ I=\int_{(u,v)\in [a,b]\times [c,d]}f(u,v)dg(u,v) =\int_{u=a}^{b}(\int_{v=c}^{d}f(u,v) dh_{2}(v)) dh_{1}(u) }$      :?:

I don't know what $\int_{(u,v)\in [a,b]\times [c,d]}f(u,v)dg(u,v)$ means. :|

Is it some kind of product measure of borel measures which are defined by integration against the $h_{1}$ and $h_{2}$? If $h_{1}$ and $h_{2}$ are appropriate kinds of functions it seems like it should make sense... but of course if I'm thinking of it correctly, the answer follows right away from the general version of fubini for abstract measures...


Solve thiss equation in R:

$ 2\sqrt {27x^2  \plus{} 24x \plus{} \frac{{28}}{3}}  \equal{} 1 \plus{} \sqrt {\frac{{27}}{2}x \plus{} 6}$

hello,  as the range of definition we have the interval $ \frac{4}{9} \le x<\infty$, after squaring both sides and simplifying we have $ 108x^2\frac{165}{2}x\plus{}\frac{91}{3}\equal{}2\sqrt{\frac{27}{2}x\plus{}6}$.
Squaring again we get $ 11664x^4\plus{}17820x^3\plus{}\frac{53433}{4}x^2\plus{}4951x\plus{}\frac{8065}{9}\equal{}0$.
This equation you must solve.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello,  as the range of definition we have the interval $ \frac {4}{9} \le x < \infty$, after squaring both sides and simplifying we have $ 108x^2\frac {165}{2}x \plus{} \frac {91}{3} \equal{} 2\sqrt {\frac {27}{2}x \plus{} 6}$.
Squaring again we get $ 11664x^4 \plus{} 17820x^3 \plus{} \frac {53433}{4}x^2 \plus{} 4951x \plus{} \frac {8065}{9} \equal{} 0$.
This equation you must solve.
Sonnhard.[/quote]
I think we can't to solve such as that, the 4th equation is used calculus to solve anh I hate it!! ^^ Can you try again?? ;)


[quote="cdymdcool"]Find the midline of y=45+18sin(2x)?[/quote]
I would assume that because a trapezoidal mid-line is a line that passes through the middle of the trapezoid (equidistant from both bases) that this is the same...

So I guess:
[hide]$y=45$[/hide]


Pišem TeX datoteku u kojoj momram istovremeno koristiti i latinicu i (našu) ćirilicu. Vidio sam na netu mnoga rješenja, ali nemam baš volje da toliko kompliciram: a datoteku svakako moram predati u TeX obliku, što znači da rješenje mog problema ne smije biti previše komplicirano, da bi uredništvo kom predajem datoteku moglo s njom uopće raditi.

Ima li iko iskustva sa sličnim situacijama?

Unaprijed zahvalan  :lol:

Pozdrav!

Mislim da sam ti nasao jednostavno rjesenje. Mislim, trebas ga probat. :)

Ima na http://www.agnld.uni-potsdam.de/~msremac/My_name/my_name.html jedan krasan source iz kojeg zakljucujem da su sljedece mogucnosti uz pretpostavku da je default pismo stavljeno [b]cirilicno[/b].

Predlazem sljedece kad zatreba latinica:
1) \textrm{tekst kojeg zelis na latinici}
2) {\rm tekst kojeg zelis na latinici}
3) napravi naredbu, predlazem: \LAT[1]{\rm {#1}} ili slicno \LAT[1]{\textrm{#1}}

Matematicki dio ionako ide na latinicnom pismu :).

Bilo bi dobro kad bi isprobao i ovdje stavio source i pdf s uspjesnim rjesenjem za buduce narastaje. Trenutno nemam nimalo vremena da pokusavam.

EDIT1: Evo i ja jos jedno pitanje. Ako postoji \textrm i \rm, zna li netko postoje li ekvivalenti za cirilicu (u slucaju da mi zatreba, ipak mi je default latinica)?

EDIT2: Ocito imam vrimena za prckanje po netu i da sam sebi odgovorim (a stoji u istom sourceu): \cyrrm i \textcyrrm :)

Hvala! Koristeći navedeni source, riješio sam $ 93.33\%$ svog problema. Naime, $ 28$ velikih i malih slova radi. No, slovo ć (ћ) se ispisuje kao ц sa kvačicom, a slovo dž (џ) se uopće ne ispisuje. No, nisam siguran da li je to lokalni problem (na mom kompjuteru), ili uopće, zato prilažem i source i kompajliranu verziju djelića dokumenta koji pišem uz inkorporiranu tabelu iz gore navedenog sourcea, tek da se vidi šta se ispisuje i kako. Kokane, ako imaš vremena da kompajliraš source kod sebe, tek da vidiš da li dobijaš iste rezultate kao i ja, bio bih ti jako zahvalan.

Sad stvarno nemam kad... :(
Ono sto vidim odoka je da ti dz ne radi jer uopce nemas definirane znakove \Dzh i \dzh.
A sto se tice ovog drugog, ne znam...

Mislim da bi bilo najbolje kad bi se javio netko iz Srbije tko inace pise na cirilici u TeX-u/LaTeX-u i dao ti svoj template, jer ocito je stvar encodinga i slicnih stvari.

U preambuli je definisano \dj i \Dj, ali ne i \Dzh i \dzh (to su 2 error-a).
Bas zbog problema sa cirilicom uvek pisem latinicom, ali postoje paketi za cirilicu, probaj sa
\newfont{\cir}{wncyr10 scaled \magstephalf} dobija se cirilica 11pt. Ovo resava problem samo za krace dokumente. Inace, ž se uvodi sa newcommand{\Z}{\symbol{'21}} , a ć se uvodi sa \newcommand{\cc}{\symbol{'17}} ako se koristi definisani font iznad.
Naravno, ne mora biti \Z i \cc
Za ostale ne znam.

Svi problemi su manje-više sad otklonjeni. Istraživao sam malo šta koji symbol predstavlja, pa sam dopunio svoju preambulu onim što mi fali:

\usepackage[OT2,OT1]{fontenc}
\def\cyr{\fontencoding{OT2}\fontfamily{wncyr}\selectfont}
\def\cyrrm{\fontencoding{OT2}\fontfamily{wncyr}\selectfont}
\def\cyrss{\fontencoding{OT2}\fontfamily{wncyss}\selectfont}
[b]\newcommand{\dzh}{\symbol{'12}}
\newcommand{\Dzh}{\symbol{'2}}
\newcommand{\cj}{\symbol{'17}}
\newcommand{\Cj}{\symbol{'7}}[/b]

Nije cirilica ali je TeX. U PDF-u u prvom zadatku na prvoj strani (podskupovi ravni), koji je kod za slova A i B (boldovana, malo cudna :lol: znate na koje mislim)?

\mathcal{A} i \mathcal{B} - kaligrafski znakovi

Da,da,da jeste. Skroz sam zaboravio... Hvala puno!

Jos jedno pitanje, u novom PDF-dokumentu, kako se pravi ''slika'' na 5. strani, 2. zadatak?

Sigurno se radi o picture environmentu. Potrazi malo po netu kako iskodirat takve slike. Naredbe put, multiput i slicno...

http://www.mathlinks.ro/viewtopic.php?t=246732

Harune, mozda ovo pomogne i bude lakse za rad ubuduce.

Definitivno pomaže (primjetio sam poslije u nekim radovima drugih kolega slična zaglavlja i konstrukcije, tako da planiram koristiti ovakav princip).

BTW, zaboravio sam ostaviti kako izgleda rezultat mog pionirskog rada u ĆiroTeXu  :lol:

http://projectpen.files.wordpress.com/2009/01/gpen-v01n1-ser-cyr.pdf

preuzeto sa http://projectpen.wordpress.com/global/

Nasao sam jos malo o temi, bez obzira sta je o ruskoj cirilici rijec:
http://www.cromwell-intl.com/russian/latex.html


limit
trigonometry
calculus
calculus computations

how do i find the limit/s        of

F(x) = 2^(-x/4)cosx^2
with x going to positive infinity
and to the negative infinity

also how would i write it in limit notation??

You don't want to ever write $ \lim_{x\to\infty}\cos x^2$ since that limit doesn't exist. This is a classic squeeze theorem case:

We have $ \minus{} 2^{ \minus{} x/4}\le F(x)\le 2^{ \minus{} x/4}.$

Since $ \lim_{x\to\infty} \minus{} 2^{ \minus{} x/4} \equal{} \lim_{x\to\infty}2^{ \minus{} x/4} \equal{} 0,$ then $ F(x)$ has that same limit as well.

I assume that this is what ibrahimsenturk thought he was saying, but it isn't what he actually wrote down - be careful of the language used.

As for the limit as $ x\to \minus{} \infty:$ not only does that limit fail to exist, but $ F(x)$ takes on every real value infinitely many times in each interval of the form $ (\infty, \minus{} M).$


show that if $a+b+c\le6$ for real numbers $a, b, c\ge0$ then
$\frac{a}{a^{2}-1}+\frac{b}{b^{2}-1}+\frac{c}{c^{2}-1}\ge2$

something is wrong here.......
take $a,b,c\leq1$, the condition holds, but the inequality doesn't.......


geometry
geometric transformation
homothety
ratio
analytic geometry
geometry proposed

Hi everyone I'm a new member in this forum. I'm dealing with a geometry problem so I need your help. Sorry if I use wrong geometrical terms. The problem is: We have a semicircle AB and K is it's centre. C is a random spot. DC is vertical to AB. AC and CB are also semicircles. ε1 and ε2 are tangents. We have to prove that the ε1 and ε2 are parallels (ε1//ε2). There is an image in order to help you. I appreciate your help.

[url=http://img265.imageshack.us/my.php?image=geometryproblemjp2.png][img]http://img265.imageshack.us/img265/9417/geometryproblemjp2.th.png[/img][/url]

Double post here: http://www.mathlinks.ro/viewtopic.php?t=155180

I think that it is not as difficult .
Let $K,\ L$ the points of tagency of the lines $e_{1}$ with the circles with diameters $AC,\ CB$ respectively .Then it is easy to find that the points
$A,K,D$ are collinear and the points $B,L,D$ are collinear .
Finally $\angle{DBA}=\angle{KCA}$ or $\angle{CKe_{1}}=\angle{ADe_{2}}$

so done  :)   .

Also a simple and short solution can be obtained by analytic geometry .

Let $\epsilon_{1}\cap CD=P$. let $\epsilon_{1}$ touch the semicircles with diameters $AC$ and $BC$ at $E$ and $F$. let $O_{1}$ and $O_{2}$ denote the centers of the semicircles. it's not difficult to see that a homothety with center $C$ and  ratio 2 sends $\triangle O_{1}PO_{2}$ to $\triangle ADB$ (sends $CO_{1}$ to $CA$ and $CO_{2}$ to $CB$). so $AD=2O_{1}E$. combining this with $AC=2O_{1}C$ we get $AD\parallel O_{1}P$. in the same way $BD\parallel O_{2}P$. now $\epsilon_{1}$ is tangent to circle $ADB$ at $D$, so $\epsilon_{2}$ is tangent to circle $O_{1}EO_{2}$ at $E$. hence by homothety again we conclude that $\epsilon_{1}\parallel \epsilon_{2}$.


if a+b+c=1, prove that

$ abc(1\minus{}c) \leq \frac{1}{4} (a\plus{}b)^3$

[quote="4865550150"]if a+b+c=1, prove that

$ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3$[/quote]

It's so easy and funny :P

\[ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3\Leftrightarrow abc\le \frac{(a\plus{}b)^2}{4}\]

which follows by $ c\le 1$ and $ ab\le \frac{(a\plus{}b)^2}{4}$  :lol:

[quote="Pain rinnegan"][quote="4865550150"]if a+b+c=1, prove that

$ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3$[/quote]

It's so easy and funny :P
\[ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3\Leftrightarrow abc\le \frac {(a \plus{} b)^2}{4}\]
which follows by $ c\le 1$ and $ ab\le \frac {(a \plus{} b)^2}{4}$  :lol:[/quote]

good solution

[quote="Dijkschneier"][quote="Pain rinnegan"][quote="4865550150"]if a+b+c=1, prove that

$ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3$[/quote]

It's so easy and funny :P
\[ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3\Leftrightarrow abc\le \frac {(a \plus{} b)^2}{4}\]
which follows by $ c\le 1$ and $ ab\le \frac {(a \plus{} b)^2}{4}$  :lol:[/quote]
Have you supposed that the numbers are positive ?  :maybe:[/quote]

They have to be. I supposed that the proposer forgot to mention this. Otherwise set $ (a,b,c)\equal{}(2,\minus{}3,2)$.

[quote="4865550150"]if a+b+c=1, prove that

$ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3$[/quote]

[quote="Pain rinnegan"]Otherwise set $ (a, b, c) \equal{} (2, \minus{} 3, 2)$ .
[/quote]

This does not check.  You will get $ 12 \le \frac { \minus{} 1}{4}$ in the original inequality.

-----------------------------------------------------------------------------------------

Don't assume the proposer meant positive, or even non-negative,
values for the variables.

You could challenge him/her with a clarification question on the parameters.

I think you didn't understand what i wanted to say.

[quote="Arrange your tan"]

Don't assume the proposer meant positive, or even non-negative,
values for the variables.
[/quote]

You must be kidding. Above i said that: if $ a,b,c\ge 0$ the problem is true, if not the problem is false. So i don't understand why you need a clarification.

[quote="Arrange your tan"]

[quote="Pain rinnegan"]Otherwise set $ (a, b, c) \equal{} (2, \minus{} 3, 2)$ .
[/quote]

This does not check.  You will get $ 12 \le \frac { \minus{} 1}{4}$ in the original inequality.

[/quote]

I think you should have understand that by $ (a,b,c) \equal{} (2, \minus{} 3,2)$ i meant a counterexample for the case when $ a,b,c$ can be negative, from which it follows that $ a,b,c$ are necessarily positives.

[quote="Arrange your tan"]

You could challenge him/her with a clarification question on the parameters.[/quote]

Sorry , but i think your advice is unuseful, because even without details the problem can be interpretated.

So here goes my advice for you: next time when you want to clarify a thing be more carefull, because now instead of clarifying it you made it more unclear.

[quote="Pain rinnegan"]I think you didn't understand what i wanted to say.

You must be kidding. Above i said that: if $ a,b,c\ge 0$ the problem is true, if not the problem is false.
So i don't understand why you need a clarification.
[/quote]

No, you did not.

[quote]
[quote"Pain rinnegan"]
I think you should have understand that by $ (a,b,c) \equal{} (2, \minus{} 3,2)$ i meant a counterexample for the case when $ a,b,c$ can be non-negative, from which it follows that $ a,b,c$ are necessarily positives.
[/quote]

No, that does not follow.  A solution is $ (0, 0, 1)$.  Non-negative includes that they can
be zero or positive.  So, for example, you are missing the solution $ (0,0,1)$.

[quote="Arrange your tan"]

You could challenge him/her with a clarification question on the parameters.[/quote]

[quote="Pain rinnegan"]
Sorry , but i think your advice is unuseful, because even without details the problem
can be interpretated.
[/quote]
Of course it's useful, unless you choose not to do it.
You interpreted the parameters wrong yourself.

[quote="Pain rinnegan"]
It's so easy and funny
[/quote]

No, those adjectives are subjective.  Simply stop using them.
You messed up in your solutions, so it wasn't "so easy" after all.

Here is a *factual* statement you could make:
"I found this problem easier than some other one."

And, by typing "so easy," "funny," and using a smiley face,
you have a bad problem thinking too highly of yourself.
Anyone with your attitude is to be knocked down to size.
Approach posts with humility.

There are an infinite number of solutions not accounted for.

Let $ k$ be any real number.

Then the solution $ (k, \minus{}k, 1)$ satisfy the inequality with the condition
placed on the variables.

[quote="geniusbliss"][quote="4865550150"]if a+b+c=1, prove that

$ abc(1 \minus{} c) \leq \frac {1}{4} (a \plus{} b)^3$[/quote]
next time please mention the constraint for the concerned variables and i dont think it can be assumed for all reals if nothing had been mentioned.[/quote]

No, it's the other way around.   The real numbers *cannot*
be discounted if it has not been mentioned.

[quote="Arrange your tan"][quote="Pain rinnegan"]I think you didn't understand what i wanted to say.

You must be kidding. Above i said that: if $ a,b,c\ge 0$ the problem is true, if not the problem is false.
So i don't understand why you need a clarification.
[/quote]

No, you did not.

...

Anyone with your attitude is to be knocked down to size.
Approach posts with humility.[/quote]

While I don't wish to belabor this thread, which was already resolved, I just wanted to point out that Pain Rinnegan was right up to semantics. His selection between "$ a,b,c \ge 0$" and "$ a,b,c \in \mathbb{R}$" was fairly reasonable underneath the context of the original problem, and you misunderstood the point he was trying to make: he showed that the latter assumption could not be that of the problem, because of the counterexample he pointed out (on the other hand, the question was not to find all possible solutions of $ (a,b,c)$). It's fairly pointless to make pedantic corrections based upon language - not everyone who posts in this forum is a native English speaker  :).

[quote="azjps"][quote="Arrange your tan"][quote="Pain rinnegan"]I think you didn't understand what i wanted to say.

You must be kidding. Above i said that: if $ a,b,c\ge 0$ the problem is true, if not the problem is false.
So i don't understand why you need a clarification.
[/quote]

No, you did not.

...
[/quote]
While I don't wish to belabor this thread, which was already resolved, I just wanted to point out that Pain Rinnegan was right up to semantics. His selection between "$ a,b,c \ge 0$" and "$ a,b,c \in \mathbb{R}$" was fairly reasonable underneath the context of the original problem, and you misunderstood the point he was trying to make: he showed that the latter assumption could not be that of the problem, because of the counterexample he pointed out (on the other hand, the question was not to find all possible solutions of $ (a,b,c)$). It's fairly pointless to make pedantic corrections based upon language - not everyone who posts in this forum is a native English speaker  :).[/quote]

Your even alleging that he was "right up to semantics" is immaterial.
Don't try to defend someone's intentions as in "he meant well, but
fell short."  You *are* belaboring this thread, so your wish not to is inconsistent.
 (In contrast, I have not made a claim to not belabor
this thread.)  Otherwise you should not have posted if it was already
resolved to you.  He didn't select $ "a, b, c \ge 0"$ first.  That claim
came later.  First, he answered in the affirmative to a poster asking
if he supposed the numbers to be positive.   That's changing his story.
No, it's wrong to back someone for assuming something here where
there's nothing about the context of the problem to steer towards
using only nonpositive integers or only nonnegative integers.   No, he
did not show the latter condition of the problem ("$ a,b,c \in \mathbb{R}$") could not be that of the problem, because I posted an answer of a set of infinite solutions where two of the variables belong to the real numbers.
It's laziness and not owning up to the correct meanings, instead of
someone such as yourself making excuses for incomplete and/or sloppy
work by casting the word "pendantic" at someone when it doesn't suit you.  Bringing up that certain others here are not native English speakers is a deflection.  He knew enough about the English posed to him, knew what
he was typing, but his logic and his recall about part of what he posted were/are not true.

[quote="Arrange your tan"][/quote][quote="azjps"][quote="Arrange your tan"][quote="Pain rinnegan"]I think you didn't understand what i wanted to say.

You must be kidding. Above i said that: if $ a,b,c\ge 0$ the problem is true, if not the problem is false.
So i don't understand why you need a clarification.
[/quote]

No, you did not.

...
[/quote]
While I don't wish to belabor this thread, which was already resolved, I just wanted to point out that Pain Rinnegan was right up to semantics. His selection between "$ a,b,c \ge 0$" and "$ a,b,c \in \mathbb{R}$" was fairly reasonable underneath the context of the original problem, and you misunderstood the point he was trying to make: he showed that the latter assumption could not be that of the problem, because of the counterexample he pointed out (on the other hand, the question was not to find all possible solutions of $ (a,b,c)$). It's fairly pointless to make pedantic corrections based upon language - not everyone who posts in this forum is a native English speaker  :).[/quote][quote="Arrange your tan"]

Your even alleging that he was "right up to semantics" is immaterial.
Don't try to defend someone's intentions as in "he meant well, but
fell short."  You *are* belaboring this thread, so your wish not to is inconsistent.
 (In contrast, I have not made a claim to not belabor
this thread.)  Otherwise you should not have posted if it was already
resolved to you.  He didn't select $ "a, b, c \ge 0"$ first.  That claim
came later.  First, he answered in the affirmative to a poster asking
if he supposed the numbers to be positive.   That's changing his story.
No, it's wrong to back someone for assuming something here where
there's nothing about the context of the problem to steer towards
using only nonpositive integers or only nonnegative integers.   No, he
did not show the latter condition of the problem ("$ a,b,c \in \mathbb{R}$") could not be that of the problem, because I posted an answer of a set of infinite solutions where two of the variables belong to the real numbers.
It's laziness and not owning up to the correct meanings, instead of
someone such as yourself making excuses for incomplete and/or sloppy
work by casting the word "pendantic" at someone when it doesn't suit you.  Bringing up that certain others here are not native English speakers is a deflection.  He knew enough about the English posed to him, knew what
he was typing, but his logic and his recall about part of what he posted were/are not true.[/quote]

You're very annoying because you speak about inequalities but you don't know anything about them.

Once again i'll tell you: the question in this thread is to prove an inequality, not solving an inequation, but you still insists with a set of infinite solutions for real numbers.

BTW you're reply is so long and boring that i prefered not reading it all. I'll repeat myself: you should write a book about fairy-tales because you seem more interested in semantics than in mathematics and you have a huge power of fascination.

But since my solution was both correct and clear i don't have why to waste my time with you.

[quote="Arrange your tan"]  No, he
did not show the latter condition of the problem ("$ a,b,c \in \mathbb{R}$") could not be that of the problem, because I posted an answer of a set of infinite solutions where two of the variables belong to the real numbers.
It's laziness and not owning up to the correct meanings, instead of
someone such as yourself making excuses for incomplete and/or sloppy
work by casting the word "pendantic" at someone when it doesn't suit you.[/quote]

By semantics, I was referring to your choice of pointing out $ (0,0,1)$ as satisfying the inequality as an example of "non-negative" versus "positive", where Pain Rinnegan clearly made a typo. I'm not saying that's a bad thing, and I apologize if 'pedantic' was too strong of a word; it just struck me as a bit pointed. Though I do think Pain Rinnegan "meant well, but he didn't fall short", to use your words.

By the way, the question is a phrased as a "prove that" question, not as a "find all solutions $ (a,b,c)$ that satisfy this inequality', which I believe you might have mistaken the question as (fortunately, that is an easy extension of the given solution  :) ).

We need a moderator to lock this thread, it is too silly.

@ Pain Rinnegan: There was definitely no need for the personal attacks, Arrange Your Tan made an interpretation mistake, and that's no big deal.