59th Ukrainian National Mathematical Olympiad

From the description of the problem we have that $xy+yz+zx=3xyz$. The given inequality is equivalent to $$\begin{gathered} (x-1)(y-1)(z-1)=(xyz-1)+(x+y+z-xy-yz-xz)=(xyz-1)+(x+y+z-3xyz)=x+y+z-1-2xyz\le \frac{1}{4}(xyz-1)\iff \\ x+y+z\le \frac{9}{4}xyz+\frac{3}{4}. \end{gathered}$$

Let us get rid of the dependence between the variables $x$, $y$, $z$, choose the positive ones $a$, $b$, $c$, for which the following equality is true: $$x=\frac{a+b+c}{3a},y=\frac{a+b+c}{3b},z=\frac{a+b+c}{3c}.$$

It is clear that for any positive $a$, $b$, $c$ the condition $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ is fulfilled, and the inequality that must be proved is rewritten as follows: $$\left(\frac{a+b+c}{3}\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le \frac{9}{4}\cdot \frac{(a+b+c{)}^3}{27abc}+\frac{3}{4}\iff 4(a+b+c)(ab+bc+ca)\le (a+b+c{)}^3+9abc.$$ After opening the brackets and reducing fraction we get: $$\begin{array}{rl}& 4(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)+12abc\le \\ & \le a^3+b^3+c^3+3(a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b)+15abc\iff \\ & a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b+3abc\le a^3+b^3+c^3,\end{array}$$

which is a case of Schur's inequality