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We have $n+1$ real numbers $x_0$ to $x_n$, with $x_0=0, x_n=1$ and $x_i < x_{i+1}$ for all $0 \le i < n$.

Prove: $$\sum_{i=0}^{n-1}\sqrt{x_{i+1}^2-x_i^2} \le \sqrt n$$

I think I have seen this problem before (it may be well known) but I'm not sure where and I couldn't find it.

I also found a geometric way to see the problem as in the following figure (taking $n=3, a=x_1, b=x_2$), the summation would be the perimeter of the figure $-1$. We can also extend the geometric representation by repeatedly adding smaller right angled triangles to the end.

Geometric Representation

Thanks for any help in advance.

Kevin Xie
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1 Answers1

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Consider the CS inequality $$1\cdot n=\sum (x_{i+1}^2-x_i^2)\cdot \sum 1 \geqslant\left(\sum\sqrt{x_{i+1}^2-x_i^2}\right)^2$$ where the sums range over $0\leqslant i <n$.

Macavity
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