63rd Czech and Slovak Mathematical Olympiad

Since the inequality involves the minimum of two positive numbers and since the function $y=x^2$ is increasing on the set ${\mathbb{R}}^{+}$, our task is to verify $$(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le {\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)}^2\text{and}(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le {\left(\frac{y}{x}+\frac{x}{z}+\frac{z}{y}\right)}^2,(1)$$

as well as to find when at least one equality in (1) holds. Replacing a triple ($x$, $y$, $z$) by the triple ($y$, $x$, $z$), we get the second inequality in (1) from the first one. Thus we can restrict to the proof of the first inequality. Distributing both sides leads to $$3+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\le \frac{x^2}{y^2}+\frac{y^2}{z^2}+\frac{z^2}{x^2}+2\left(\frac{x}{z}+\frac{y}{x}+\frac{z}{y}\right).$$ Let us introduce the new (positive) variables $a=x/y$, $b=y/z$, $c=z/x$ and rewrite the last inequality as $$\left(a^2-1-a+\frac{1}{a}\right)+\left(b^2-1-b+\frac{1}{b}\right)+\left(c^2-1-c+\frac{1}{c}\right)\ge 0.(2)$$ For any positive $t$, we notice that $$t^2-1-t+\frac{1}{t}=(t^2-1)-\frac{t^2-1}{t}=\frac{(t^2-1)(t-1)}{t}=\frac{(t-1{)}^2(t+1)}{t}.$$ This implies that (2) holds as well and that (2) becomes an equality if and only if $a=b=c=1$, i.e. $x=y=z$ for the original variables. Note that the last condition does not change under transformation $(x,y,z)\to (y,x,z)$. Thus the original inequality is proven and becomes an equality if and only $x=y=z$.