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