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Let $a,b,c$ are real number such that $a+b+c=1$. Prove that: $$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.$$

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Where did you see this problem? What have you tried? – Jesse Madnick Dec 29 '12 at 6:41
it 's similar as Poland MO , but it 's harder, Poland contest give us $a,b,c \geq \frac{-3}{4}$ but this inequality 's also true for real number – Haruboy15 Dec 29 '12 at 6:43
Haruboy: How did you solve the version from the Poland contest? – Jonas Meyer Dec 29 '12 at 6:46
Base on tangent method. We prove that : $\frac{a}{a^2+1} \leq \frac{18}{25}(a-\frac{1}{3})+\frac{3}{10}$ $\Leftrightarrow (a-\frac{1}{3})^2(4a+3) \geq 0$ It's true. and done – Haruboy15 Dec 29 '12 at 6:54
@Haruboy15 This inequality only holds when $a>-4/3$ in Poland contest,but this time there is no such restriction – Golbez Dec 29 '12 at 7:23
up vote 1 down vote accepted

As you have known, the inequality holds when $a,b,c\ge -\frac{3}{4}$. To complete the proof, without loss of generality, we may assume that $a\le -\frac{3}{4}$.

Note that $\frac{b}{b^2+1},\frac{c}{c^2+1}\le \frac{1}{2}$. Since when $-9\le a\le -\frac{3}{4}$, $\frac{a}{a^2+1}<-\frac{1}{10}$, the inequality holds when $a\ge -9$. When $a\le-9$, $b+c\ge 10$. Without loss of generality, we may assume that $b\ge 5$. Then $\frac{b}{b^2+1}\le \frac{5}{26}<\frac{2}{5}$, so the inequality also holds.

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Let $S(a,b,c) = \frac{a}{a^2+1}+ \frac{b}{b^2+1}+ \frac{c}{c^2+1}$. Since we have that $a,b$ and $c$ are all represented symmetrically in $S$ (interchanging any two variables has no effect), it must be that case that $S$ has a stiationary value (maximum or minimum) when all parameters are equal, ie. $a=b=c=x$. Using the fact that $a+b+c=1$ gives $x=\frac{1}{3}$, so we must classify the point $(a,b,c)=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ as a maximum or minimum of the function $S(a,b,c)$. Consider increasing the value of an parameter by a small quantity $\delta$, and thereby decreasing the value of another parameter by $\delta$. In other words, without loss of generality take $a=\frac{1}{3}+\delta$, $b=\frac{1}{3}-\delta$ and $c=1/3$. We then have that:

$$ S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})=\frac{3}{10}+\frac{\frac{1}{3}+\delta}{(\frac{1}{3}+\delta)^2+1} + \frac{\frac{1}{3}-\delta}{(\frac{1}{3}-\delta)^2+1} $$ $$ S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})\approx\frac{3}{10}+\frac{\frac{1}{3}+\delta}{\frac{10}{9}+\frac{2\delta}{3}} + \frac{\frac{1}{3}-\delta}{\frac{10}{9}-\frac{2\delta}{3}}=\frac{33}{10}+\frac{6}{3\delta-5}-\frac{6}{3\delta+5} $$

Then, to leading order in $\delta$, we have the following series expansion:

$$ S(\frac{1}{3}+\delta,\frac{1}{3}-\delta,\frac{1}{3})\approx \frac{9}{10} - \frac{108}{125}\delta^2 $$

Thus, $\delta=0$ is a maximum, which tells us that $(a,b,c)=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is a maximum! Furthermore, this tells us that $S_{\text{max}} = \frac{9}{10}$ which immediately implies that $S(a,b,c)\leq \frac{9}{10}$ as required.

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Could you expand on why symmetry gives us a stationary value at $(1/3,1/3,1/3)$? – Andrés E. Caicedo Dec 29 '12 at 16:56
@ Andres Caicedo It's something of an appeal to the principle of stationary action: This problem has a physical analogue (excuse my bad drawing): And then clearly the question is to show that $\sin(\alpha)+\sin(\beta)+\sin(\gamma)\leq\frac{9}{10}$. This is where the symmetry part comes into play, in essence: why would the universe "favor" one triangle over another when their contributions to $S$ are identical? – Mr. G Dec 30 '12 at 3:59
Sorry to be a bother, but would you mind walking me through the formal argument in this case? – Andrés E. Caicedo Dec 30 '12 at 4:19
@Adres Caicedo Ah yes, a formal argument. Er well, take $\epsilon>0$ and then ... well ... um ... honestly, I don't have a formal argument. The PoSA has mathematical justification in the form of the Euler-Lagrange equations, but these return functions rather than values so this isn't the greatest problem to actually use it. However, the spirit of the PoSA still applies in the form of the "nature puts in the least amount of effort possible" argument that physicists are always hoping they can invoke. I understand that a mathematician probably won't find this compelling at all, sorry :( – Mr. G Dec 30 '12 at 7:09

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