Let $k \geqslant 0$ be given. Determine the minimum of the expression $A=\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}}+\frac{z}{1+t^{2}}+$ $\frac{t}{1+z^{2}}$ under the condition $x+y+z+t=k$ and $x, y, z, t$ positive.
We apply $\left(\frac{x}{X}+\frac{y}{Y}\right)(x X+y Y) \geqslant(x+y)^{2}$ to $X=1+y^{2}$ and $Y=1+x^{2}$, which gives $\left(\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}}\right)((x+y)(1+x y)) \geqslant(x+y)^{2}$. If $(x, y) \neq(0,0)$, this simplifies to $\frac{x}{1+y^{2}}+$ $\frac{y}{1+x^{2}} \geqslant \frac{x+y}{1+x y^{2}}$, an inequality that is always true for $x=y=0$. Since $x y \leqslant(x+y)^{2} / 4 \leqslant k^{2} / 4$, we deduce that $\frac{x}{1+y^{2}}+\frac{y}{1+x^{2}} \geqslant \frac{4(x+y)}{4+k^{2}}$, hence $A \geqslant \frac{4 k}{4+k^{2}}$, a bound that is achieved for $x=y=k / 2$ and $z=t=0$ for example.