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 ?

Let $x=\frac{a}{3}$, $y=\frac{b}{3}$ and $z=\frac{c}{3}$.
Hence, $a+b+c=3$ and we need to prove that $$\sum_{cyc}\frac{a}{9+a^2}\leq\frac{3}{10}$$ or
$$\sum_{cyc}\left(\frac{1}{10}-\frac{a}{9+a^2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a^2-10a+9}{9+a^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{a^2-10a+9}{9+a^2}+\frac{2}{3}\right)\geq2$$ or
$$\sum_{cyc}\frac{(a-3)^2}{9+a^2}\geq\frac{6}{5}.$$
Let $k=3\sqrt2-6$. Hence, by C-S
$$\sum_{cyc}\frac{(a-3)^2}{9+a^2}=\sum_{cyc}\frac{(a-3)^2(a+k)^2}{(a+k)^2(9+a^2)}\geq\frac{\left(\sum\limits_{cyc}(a-3)(a+k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}=\frac{\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2}{\sum\limits_{cyc}(a+k)^2(a^2+9)}.$$
Thus, it remains to prove that
$$5\left(\sum\limits_{cyc}(a^2-3-2k)\right)^2\geq6\sum\limits_{cyc}(a+k)^2(a^2+9),$$
which is fourth degree.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative and $abc=w^3$.
Hence, the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove the last inequality for en extremal value of $w^3$, 
which happens for equality case of two variables.
Let $b=a$ and $c=3-2a$. We obtain
$$(a-1)^2(\sqrt2a-3\sqrt2+3)^2\geq0.$$
Done!
