UkraineMO

Firstly, we will prove that if $\alpha$, $\beta \in [0, 1]$, then $\alpha^4 + \beta^5 + (\alpha - \beta)^6 \le 2$. Indeed, $t^n \le t$ for all real $t \in [0, 1]$ and any positive integer $n$. Suppose that $\alpha \ge \beta$ (case $\alpha \le \beta$ is considered analogously). Then $0 \le \alpha - \beta \le 1$, so $(\alpha - \beta)^6 \le \alpha - \beta$. Also, $\alpha^4 \le \alpha$ and $\beta^5 \le \beta$. Summing up the previous inequalities we obtain:
$$\alpha^4 + \beta^5 + (\alpha - \beta)^6 \le \alpha + \beta + \alpha - \beta = 2\alpha \le 2.$$
Now let $x$, $y$, $z \in [0, 1]$. By the latter claim:
$$x^4 + y^5 + (x-y)^6 \le 2, \quad y^4 + z^5 + (y-z)^6 \le 2, \quad z^4 + x^5 + (z-x)^6 \le 2.$$
After summing up we get the desired inequality.