62nd Belarusian Mathematical Olympiad

We use weighted Chebyshev's inequality: if $a_1,\dots ,a_n>0$, and $x_1\le x_2\le \cdots \le x_n$, $y_1\le y_2\le \cdots \le y_n$, then $$\frac{(a_1{x}_1+\cdots +a_n{x}_n)(a_1{y}_1+\cdots +a_n{y}_n)}{a_1+\cdots +a_n}\le a_1{x}_1{y}_1+\cdots +a_n{x}_n{y}_n.(\ast )$$

Now let $p(x)=a_1+a_{2}x+\cdots +a_n{x}^{n-1}$ be the given polynomial. Then we simply set in $(\ast )$ $x_1=1$, $x_2=x$, ..., $x_i=x^{i-1}$, $y_1=1$, ..., $y_i=y^{i-1}$, thus finishing the proof.