<s>[INST] 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:
$$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.$$
 [/INST] 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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