Home » Posts tagged 'What’s New?'

Tag Archives: What’s New?

A College Level Problem on Jensen’s Inequality

Saturday, March 12th, 2011 08:47 / 1 Comment

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

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