58th Ukrainian National Mathematical Olympiad

We make a substitution: $a=\frac{2x}{2x+y}$, $b=\frac{2y}{2y+z}$, $c=\frac{2z}{2z+x}$, then the inequality will be as follows: $$2\cdot {\left(\frac{x}{2x+y}\right)}^2+2\cdot {\left(\frac{y}{2y+z}\right)}^2+2\cdot {\left(\frac{z}{2z+x}\right)}^2+\frac{9xyz}{(2x+y)(2y+z)(2z+x)}=\frac{a^2+b^2+c^2}{2}+\frac{9}{8}abc\le 1,\text{ or}$$ $$a^2+b^2+c^2+\frac{9}{4}abc\le 2.$$

Consider the expression: $$\begin{array}{rl}\left(\frac{1}{a}-1\right)\left(\frac{1}{b}-1\right)\left(\frac{1}{c}-1\right)& =\left(\frac{2x+y}{2x}-1\right)\left(\frac{2y+z}{2y}-1\right)\left(\frac{2z+x}{2z}-1\right)=\frac{y}{2x}\cdot \frac{z}{2y}\cdot \frac{x}{2z}=\frac{1}{8},\text{ or}\\ \frac{(1-a)}{a}\cdot \frac{(1-b)}{b}\cdot \frac{(1-c)}{c}& =\frac{1}{8}\iff 8(1-a)(1-b)(1-c)=abc\iff 8(1-a)(1-b)(1-c)=abc\\ & \iff 8-8(a+b+c)+8(ab+bc+ca)-8abc=abc\Rightarrow \\ & \frac{9}{4}abc=2-2(a+b+c)+2(ab+bc+ca).\end{array}$$ Continue the proof using the obtained conditions: $$\begin{array}{rl}& a^2+b^2+c^2+\frac{9}{4}abc-2=\\ & =(a+b+c{)}^2-2(ab+bc+ca)+2-2(a+b+c)+2(ab+bc+ca)-2=\\ & =(a+b+c{)}^2-2(a+b+c)\le 0.\end{array}$$

The last inequality holds when $a+b+c\le 2$.

Perform transformations: $$\begin{array}{rl}a+b+c& =\frac{2x}{2x+y}+\frac{2y}{2y+z}+\frac{2z}{2z+x}=1-\frac{y}{2x+y}+1-\frac{z}{2y+z}+1-\frac{x}{2z+x}=2+\left(1-\frac{y}{2x+y}-\frac{z}{2y+z}-\frac{x}{2z+x}\right)\le 2\\ & \iff \frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\ge 1.\end{array}$$

Prove the last inequality using Cauchy–Schwarz inequality: $$\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)(x+y+z{)}^2=\left(\frac{y}{2x+y}+\frac{z}{2y+z}+\frac{x}{2z+x}\right)(y(2x+y)+z(2y+z)+x(2z+x))\ge$$ $$\ge {\left(\sqrt{\frac{y}{2x+y}}\cdot \sqrt{y(2x+y)}+\sqrt{\frac{z}{2y+z}}\cdot \sqrt{z(2y+z)}+\sqrt{\frac{x}{2z+x}}\cdot \sqrt{x(2z+x)}\right)}^2=(x+y+z{)}^2,$$ Q.E.D.