<s>[INST] Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$. 
Find the maximum value of the expression :
  $${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$$
  where $x,y,z$ are real numbers satisfying the condition that $x+y+z=1$.

Taking $x=y=z=\frac {1}{3}$, I get the expression as $\frac {3x}{1+x^2}$, which is equal to $\frac {1}{1+{\frac{1}{9}}}$ or $\frac {9}{10}$.
How can I actually solve the problem without making unnecessary assumptions ?
 [/INST] Let $x+y+z=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
We need to prove that $$\sum_{cyc}\frac{x}{x^2+(x+y+z)^2}\leq\frac{9}{10(x+y+z)}$$ or
$$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq0,$$
which is $9w^6+A(u,v^2)w^3+B(u,v^2)\geq0$, which says that
$$9\prod_{cyc}(x^2+(x+y+z)^2)-10(x+y+z)\sum_{cyc}x(y^2+(x+y+z)^2)(z^2+(x+y+z)^2)\geq$$
$$\geq\frac{1}{81}\left(\sum_{cyc}(4x^3-3x^2y-3x^2z+2xyz)\right)^2$$
is a linear inequality of $w^3$.
Thus, by $uvw$ it remains to prove the last inequality for $y=z=1$, which gives
$$(x-1)^2(x+2)^2(79x^2+253x+253)\geq0,$$
which is obviously true.
Done!
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