Baltic Way 2025 · Problem 5

For convenience, set $m=x_1$ and $M=x_n$. For every $1\le i\le n$ we have

$$M^2-x_i^2\ge0 \qquad\text{and}\qquad x_i-m\ge0.$$

Multiplying these inequalities gives

$$(M^2-x_i^2)(x_i-m)\ge0,$$

or equivalently

$$M^2x_i+mx_i^2\ge M^2m+x_i^3.\tag{1}$$

Similarly, from $M-x_i\ge0$ and $x_i^2-m^2\ge0$ we get

$$(M-x_i)(x_i^2-m^2)\ge0,$$

which is equivalent to

$$Mx_i^2+m^2x_i\ge Mm^2+x_i^3.\tag{2}$$

Adding (1) and (2),

$$(M^2+m^2)x_i+(M+m)x_i^2\ge(M+m)Mm+2x_i^3.$$

Summing this inequality for $i=1,\ldots,n$ gives

$$\begin{aligned} &(M^2+m^2)(x_1+\cdots+x_n)+(M+m)(x_1^2+\cdots+x_n^2)\\ &\qquad\ge (M+m)Mmn+2(x_1^3+\cdots+x_n^3). \end{aligned}$$

Multiplying the given condition by $2(x_1+\cdots+x_n)$ yields

$$2(x_1^3+\cdots+x_n^3) =(x_1^2+x_n^2)(x_1+\cdots+x_n) =(m^2+M^2)(x_1+\cdots+x_n).$$

Using this identity in the preceding inequality, we obtain

$$(M+m)(x_1^2+\cdots+x_n^2)\ge(M+m)Mmn.$$

Since $M+m>0$,

$$x_1^2+x_2^2+\cdots+x_n^2\ge Mmn=nx_1x_n,$$

as required.