$\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2.$ Prove $\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2} \leq \sqrt{2}.$

Cauchy-Schwartz ... \begin{eqnarray*} \left( \frac{a}{\sqrt{1+a^2}} \frac{1}{\sqrt{1+a^2}} + \frac{b}{\sqrt{1+b^2}} \frac{1}{\sqrt{1+b^2}} + \frac{c}{\sqrt{1+c^2}} \frac{1}{\sqrt{1+c^2}} \right)^2 \\\leq \left( \frac{a^2}{1+a^2} + \frac{b^2}{1+b^2} +\frac{c^2}{1+c^2} \right) \left( \frac{1}{1+a^2} + \frac{1}{1+b^2} +\frac{1}{1+c^2} \right)= 2. \end{eqnarray*}

Let $$B:=\frac{1} {1+a^2} + \frac{1} {1+b^2} + \frac{1} {1+c^2}$$

From: $$A:=\frac{a^2} {1+a^2} + \frac{b^2} {1+b^2} + \frac{c^2} {1+c^2} = 2$$

we get $A+B =3$ so $B =1$.

Now by Cauchy inequality we have $$A\cdot B \geq \big(\underbrace{\frac{a} {1+a^2} + \frac{b} {1+b^2} + \frac{c} {1+c^2}}_{C}\big)^2$$

So we have $C^2\leq 2$ and we are done.