, 2008

We prove the inequality by induction. Obviously, for $n=1$ the equality holds and for $n=2$ we have
$$\sqrt{x_1^2 + 1} + 2\sqrt{x_2^2 + 1} \ge \sqrt{(x_1 + 2x_2)^2 + 9}.$$
After squaring both sides we rewrite it as
$$\sqrt{(x_1^2 + 1)(x_2^2 + 1)} \ge x_1 x_2 + 1.$$
If the right-hand side is negative, then the inequality holds, if not, we can square both sides and transform it into $(x_1 - x_2)^2 \ge 0$, which is true. So, for $n=2$ the inequality holds.

Let us assume the inequality holds for $n$ and prove it for $n+1$. By induction hypothesis we have
$$\sqrt{x_1^2+1} + 2\sqrt{x_2^2+1} + \dots + n\sqrt{x_n^2+1} \ge \sqrt{(x_1+2x_2+\dots+nx_n)^2 + \frac{n^2(n+1)^2}{4}},$$
so it suffices to show that
$$\begin{aligned} & \sqrt{(x_1 + 2x_2 + \dots + nx_n)^2 + \frac{n^2(n+1)^2}{4}} + (n+1)\sqrt{x_{n+1}^2 + 1} \ge \\ & \ge \sqrt{(x_1 + 2x_2 + \dots + nx_n + (n+1)x_{n+1})^2 + \frac{(n+1)^2(n+2)^2}{4}}. \end{aligned}$$
This is equivalent to
$$\begin{aligned} & (x_1 + 2x_2 + \dots + nx_n)^2 + \frac{n^2(n+1)^2}{4} + (n+1)^2 x_{n+1}^2 + (n+1)^2 + \\ & + 2(n+1)\sqrt{(x_1 + 2x_2 + \dots + nx_n)^2 + \frac{n^2(n+1)^2}{4}} \sqrt{x_{n+1}^2 + 1} \ge \\ & \ge (x_1 + 2x_2 + \dots + nx_n)^2 + (n+1)^2 x_{n+1}^2 + 2(n+1)x_{n+1}(x_1 + 2x_2 + \dots + nx_n) + \\ & \qquad + \frac{(n+1)^2(n+2)^2}{4}, \end{aligned}$$
and further to
$$2\sqrt{(x_1 + 2x_2 + \dots + nx_n)^2 + \frac{n^2(n+1)^2}{4}} \sqrt{x_{n+1}^2 + 1} \ge 2x_{n+1}(x_1 + 2x_2 + \dots + nx_n) + n(n+1).$$
If the right-hand side is negative, then the inequality holds. If not, we can square both sides and we get $4(x_1 + 2x_2 + \dots + nx_n)^2 x_{n+1}^2 + 4(x_1 + 2x_2 + \dots + nx_n)^2 + n^2(n+1)^2 x_{n+1}^2 + n^2(n+1)^2 \ge 4x_{n+1}^2(x_1 + 2x_2 + \dots + nx_n)^2 + n^2(n+1)^2 + 4n(n+1)x_{n+1}(x_1 + 2x_2 + \dots + nx_n)$ and finally
$$(2(x_1 + 2x_2 + \dots + nx_n) - n(n+1)x_{n+1})^2 \ge 0.$$
This inequality obviously holds, and the equality holds when
$$x_{n+1} = \frac{2}{n(n+1)}(x_1 + 2x_2 + \dots + nx_n).$$
By induction, the equality holds when $x_1 = x_2 = \dots = x_n = x_{n+1}$.