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## Problem

Given, $-1 \le a_1 \le a_2 \le ... \le a_n \le 1$.

Prove that

$\mathbf {\sum_{i=1}^{n-1}}$ $\sqrt {1-a_i a_{i+1} - \sqrt {(1-a_i^2) (1-a_{i+1}^2)}}$ $< \frac {\pi \sqrt {2}} {2}$.

## Solution

It is natural to make trigonometric substitution $a_i= \cos{x_i}$ for some $x_i \in [0, \pi]$, $i=1,2,. . . . . ,n.$

Note that the monotonicity of the cosine function combined with the given inequalities shows that the $x_i's$ form a decreasing sequence. The expression on the left

$\mathbf {\sum_{i=1}^{n-1}}$ $\sqrt {1-a_ia_{i+1} - \sqrt {(1-a_i^2) (1-a_{i+1}^2)}}$

$= \mathbf {\sum_{i=1}^{n-1}} \sqrt {1- \cos {x_i} \cos {x_{i+1}} - \sin {x_i} \sin {x_{i+1} } }$

$= \mathbf {\sum_{i=1}^{n-1}} \sqrt {1- \cos {(x_{i+1}-x_i)}}$

$= \sqrt{2} \mathbf {\sum_{i=1}^{n-1}} \sin {\frac {x_{i+1}-x_i} {2}}$

Here we used a subtraction and a double-angle formula. The sine function is concave down on $[0, \pi]$; hence we can use Jensen’s Inequality to obtain

$\frac {1}{n-1} \mathbf {\sum_{i=1}^{n-1}}$ $\sin {\frac {x_{i+1}-x_i} {2} }$ $\le \sin {(\frac {1}{n-1} \mathbf {\sum_{i=1}^{n-1}} \frac {x_{i+1}-x_i} {2} )}$

Hence,

$\sqrt 2 \mathbf {\sum_{i=1}^{n-1}}$ $\sin {\frac {x_{i+1}-x_i} {2} }$ $\le (n-1) \sqrt{2} \sin {\frac {x_n-x_i}{2(n-1)}}$

or,

$\sqrt {2} \mathbf {\sum_{i=1}^{n-1}}$ $\sin {\frac {x_{i+1}-x_i} {2} }$ $\le \sqrt {2} (n-1) \sin {\frac {\pi}{2(n-1)}}$

Since,

$x_n-x_i \in (0,\pi)$

Using the fact that $\sin x < x$ for all $x > 0$ yields

$\frac {\sqrt{2}(n-1) \sin \pi} {2(n-1)} \le \frac {\sqrt{2} \pi}{2}$

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What’s New ? (terrytao.wordpress.com)