Balti Tee 2020 · Ülesanne 2

Denote $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$. Then

$$\frac{1}{a \sqrt{c^{2}+1}}=\frac{1}{\frac{x}{y} \sqrt{\frac{z^{2}}{x^{2}}+1}}=\frac{y}{\sqrt{z^{2}+x^{2}}} \geqslant \frac{2 y^{2}}{x^{2}+y^{2}+z^{2}}$$

where the last inequality follows from the AM-GM inequality

$$y \sqrt{x^{2}+z^{2}} \leqslant \frac{y^{2}+\left(x^{2}+z^{2}\right)}{2} .$$

If we do the same estimation also for the two other terms of the original inequality then we get

$$\frac{1}{a \sqrt{c^{2}+1}}+\frac{1}{b \sqrt{a^{2}}+1}+\frac{1}{c \sqrt{b^{2}+1}} \geqslant \frac{2 y^{2}}{x^{2}+y^{2}+z^{2}}+\frac{2 z^{2}}{x^{2}+y^{2}+z^{2}}+\frac{2 x^{2}}{x^{2}+y^{2}+z^{2}}=2 .$$

Equality holds only if $y^{2}=x^{2}+z^{2}, z^{2}=x^{2}+y^{2}$ and $x^{2}=y^{2}+z^{2}$ what is impossible.