Mediterranean Mathematical Competition

Given inequality is equivalent to: $$\frac{1}{1+\frac{x}{y}+\frac{y}{x}}+\frac{1}{1+\frac{y}{z}+\frac{z}{y}}+\frac{1}{1+\frac{z}{x}+\frac{x}{z}}\le \frac{1}{2+\frac{z}{x}}+\frac{1}{2+\frac{x}{y}}+\frac{1}{2+\frac{y}{z}}.$$ Now we can take substitution $\frac{x}{y}=a$, $\frac{y}{z}=b$, $\frac{z}{x}=c\Rightarrow abc=1$, so our inequality becomes:

$$\frac{1}{1+a+\frac{1}{a}}+\frac{1}{1+b+\frac{1}{b}}+\frac{1}{1+c+\frac{1}{c}}\le \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c}.$$ After some computations and also using $abc=1$, this becomes equivalent to: $$\frac{3(a+b+c)+3(ab+bc+ca)+(ab+bc+ca)(a+b+c)}{(ab+bc+ca{)}^2+(ab+bc+ca)(a+b+c)+(a+b+c{)}^2}\le \frac{12+4(a+b+c)+(ab+bc+ca)}{9+2(ab+bc+ca)+4(a+b+c)}.$$ Now, we take substitution $a+b+c=S$, $ab+bc+ca=P$ and inequality becomes: $$\frac{3S+3P+SP}{P^2+PS+S^2}\le \frac{12+4S+P}{9+2P+4S}\iff P^3+4S^3+3P^{2}S+P{S}^2+6P^2\ge 27S+27P+15PS(\ast )$$ It is not difficult to prove that $S^2\ge 3P$ and also $S\ge 3,P\ge 3$ (by AM $\ge$ GM and $abc=1$). Therefore: $$\begin{array}{rl}4S^3& =4S^2\cdot S\ge 12PS,P{S}^2=S\cdot PS\ge 3PS,3P^{2}S\ge 3\cdot 3^2\cdot S=27S,\\ P^3& =P^2\cdot P\ge 3^2\cdot P=9P,6P^2=6P\cdot P\ge 6\cdot 3P=18P.\end{array}$$ By summing these inequalities we found (*) to be true so our proof is finished. Equality is obviously achieved when $$S^2=3P=9\Rightarrow a=b=c=1\Rightarrow x=y=z.$$