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Thesis: $$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1}\leq 9/10$$ Assumptions: $$a+b+c=1$$ $$ a,b,c \geq -\frac{3}{4}$$ Can someone give me a hint? I suppose there is a tricky way, not using derivatives of the function of two variables

Michael Rozenberg
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Ryszard Eggink
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    I think I asked this on AoPS a long time ago, and the hint was Cauchy-Schwarz. It may have been the Titu form of CS. – Sean Roberson Jul 05 '17 at 20:35
  • Here https://math.stackexchange.com/questions/2095843/find-the-maximum-value-of-the-expression the same inequality is shown without assuming the lower bound on $a,b,c$. – Martin R Jul 05 '17 at 21:54

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Let $a=\frac{x}{3}$, $b=\frac{y}{3}$ and $c=\frac{z}{3}$.

Hence, $x+y+z=3$, $\{x,y,z\}\subset[-\frac{9}{4},+\infty)$ and we need to prove that $$\sum_{cyc}\frac{x}{x^2+9}\leq\frac{3}{10}$$ or $$\sum_{cyc}\left(\frac{1}{10}-\frac{x}{x^2+9}\right)\geq0$$ or $$\sum_{cyc}\left(\frac{(x-1)(x-9)}{x^2+9}+\frac{4}{5}(x-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(x-1)^2(4x-9)}{x^2+9}\geq0.$$ Done!

Michael Rozenberg
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