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A College Level Problem on Jensen’s Inequality

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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1 Comment

  1. Third Grade Teacher says:
    Sunday, March 13th, 2011 at 03:36

    Hello, you might want to use “\sin” and “\cos” when typing your formulae.

    Reply

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