Slovenija 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 $$\sqrt{x_1^2+1}+2\sqrt{x_2^2+1}+\cdots +n\sqrt{x_n^2+1}\ge \sqrt{(x_1+2x_2+\cdots +n{x}_n{)}^2+\frac{n^2(n+1{)}^2}{4}},$$ so it suffices to show that $$\begin{array}{rl}& \sqrt{(x_1+2x_2+\cdots +n{x}_n{)}^2+\frac{n^2(n+1{)}^2}{4}}+(n+1)\sqrt{x_{n+1}^2+1}\ge \\ & \ge \sqrt{(x_1+2x_2+\cdots +n{x}_n+(n+1)x_{n+1}{)}^2+\frac{(n+1{)}^2(n+2{)}^2}{4}}.\end{array}$$ This is equivalent to $$\begin{array}{rl}& (x_1+2x_2+\cdots +n{x}_n{)}^2+\frac{n^2(n+1{)}^2}{4}+(n+1{)}^2{x}_{n+1}^2+(n+1{)}^2+\\ & +2(n+1)\sqrt{(x_1+2x_2+\cdots +n{x}_n{)}^2+\frac{n^2(n+1{)}^2}{4}}\sqrt{x_{n+1}^2+1}\ge \\ & \ge (x_1+2x_2+\cdots +n{x}_n{)}^2+(n+1{)}^2{x}_{n+1}^2+2(n+1)x_{n+1}(x_1+2x_2+\cdots +n{x}_n)+\\ & +\frac{(n+1{)}^2(n+2{)}^2}{4},\end{array}$$ and further to $$2\sqrt{(x_1+2x_2+\cdots +n{x}_n{)}^2+\frac{n^2(n+1{)}^2}{4}}\sqrt{x_{n+1}^2+1}\ge 2x_{n+1}(x_1+2x_2+\cdots +n{x}_n)+n(n+1).$$ If the right-hand side is negative, then the inequality holds. If not, we can square both sides and we get $4(x_1+2x_2+\cdots +n{x}_n{)}^2{x}_{n+1}^2+4(x_1+2x_2+\cdots +n{x}_n{)}^2+n^2(n+1{)}^2{x}_{n+1}^2+n^2(n+1{)}^2\ge 4x_{n+1}^2(x_1+2x_2+\cdots +n{x}_n{)}^2+n^2(n+1{)}^2+4n(n+1)x_{n+1}(x_1+2x_2+\cdots +n{x}_n)$ and finally $$(2(x_1+2x_2+\cdots +n{x}_n)-n(n+1)x_{n+1}{)}^2\ge 0.$$ This inequality obviously holds, and the equality holds when $$x_{n+1}=\frac{2}{n(n+1)}(x_1+2x_2+\cdots +n{x}_n).$$ By induction, the equality holds when $x_1=x_2=\cdots =x_n=x_{n+1}$.