Belarus — Final Round

Answer: $\frac{a+b}{2} \ge \frac{x+y}{2}$.
We prove that
$$\frac{a+b}{2} \ge \frac{x+y}{2}.$$
(*)
Indeed,
$$\frac{a+b}{2} \ge \frac{x+y}{2} \Leftrightarrow a+b \ge \sqrt{ab} + \sqrt{\frac{a^2+b^2}{2}} \Leftrightarrow$$
$$\Leftrightarrow (a+b)^2 \ge ab + \frac{a^2+b^2}{2} + 2\sqrt{ab \cdot \frac{a^2+b^2}{2}} \Leftrightarrow$$
$$\Leftrightarrow (a+b)^2 \ge 4\sqrt{ab \cdot \frac{a^2+b^2}{2}} \Leftrightarrow (a+b)^4 \ge 8ab(a^2+b^2) \Leftrightarrow$$
$$\Leftrightarrow \left(\frac{(a+b)^2}{ab}\right)^2 \ge \frac{8(a^2+b^2)}{ab} \Leftrightarrow \left(\frac{a}{b} + \frac{b}{a} + 2\right)^2 \ge 8 \cdot \left(\frac{a}{b} + \frac{b}{a}\right).$$
Let $t = \frac{a}{b} + \frac{b}{a}$; then
$$\left(\frac{a}{b} + \frac{b}{a} + 2\right)^2 \ge 8 \cdot \left(\frac{a}{b} + \frac{b}{a}\right) \Leftrightarrow (t+2)^2 \ge 8t \Leftrightarrow (t-2)^2 \ge 0.$$
Note that the equality occurs in (*) if $a = b$.