The South African Mathematical Olympiad Third Round

Note that for any $k\le n$, we have $$k{x}_k^2\le x_1^2+x_2^2+\cdots +x_k^2\le 1$$ by the given conditions. This implies that $x_k\le \frac{1}{\sqrt{k}}$ and thus $x_k^2\le \frac{x_k}{\sqrt{k}}$. We conclude that $$\frac{x_1}{\sqrt{1}}+\frac{x_2}{\sqrt{2}}+\cdots +\frac{x_n}{\sqrt{n}}\ge x_1^2+x_2^2+\cdots +x_n^2=1,$$ which proves the statement.

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Alternative solution.

We prove the inequality by induction on $n$: for $n=1$, we have $x_1=1$ by the given conditions, and there is nothing to prove. Now assume that the inequality holds for $n-1$, and consider real numbers $x_1,x_2,\dots ,x_n$ that satisfy the given conditions. As in the first solution, we find that $x_n\le \frac{1}{\sqrt{n}}$. Now set $$y_k=\frac{x_k}{\sqrt{1-x_n^2}}.$$ Then $y_1\ge y_2\ge \cdots \ge y_{n-1}$ and $$y_1^2+y_2^2+\cdots +y_{n-1}^2=\frac{1}{1-x_n^2}(x_1^2+x_2^2+\cdots +x_{n-1}^2)=\frac{1}{1-x_n^2}(1-x_n^2)=1,$$ so that we can apply the induction hypothesis to $y_1,y_2,\dots ,y_{n-1}$. We conclude that $$\begin{array}{rl}\frac{x_1}{\sqrt{1}}+\frac{x_2}{\sqrt{2}}+\cdots +\frac{x_{n-1}}{\sqrt{n-1}}+\frac{x_n}{\sqrt{n}}& =\sqrt{1-x_n^2}\left(\frac{y_1}{\sqrt{1}}+\frac{y_2}{\sqrt{2}}+\cdots +\frac{y_{n-1}}{\sqrt{n-1}}\right)+\frac{x_n}{\sqrt{n}}\\ & \ge \sqrt{1-x_n^2}+\frac{x_n}{\sqrt{n}}.\end{array}$$ It remains to show that the last expression is $\ge 1$ to complete the induction. However, this is equivalent to $$\sqrt{1-x_n^2}\ge 1-\frac{x_n}{\sqrt{n}}$$ or $$1-x_n^2\ge 1-\frac{2x_n}{\sqrt{n}}+\frac{x_n^2}{n},$$ which finally simplifies to $$x_n\left(\frac{2}{\sqrt{n}}-\left(1+\frac{1}{n}\right)x_n\right)\ge 0.$$ Since we know that $x_n\le \frac{1}{\sqrt{n}}$, we have $$\frac{2}{\sqrt{n}}-\left(1+\frac{1}{n}\right)x_n\ge \frac{2}{\sqrt{n}}-2x_n\ge 0,$$ so that both factors on the left hand side are nonnegative. This completes our proof.