## Jensen’s inequality and trigonometry

The problem of maximizing $\cos{A}+\cos{B}+\cos{C}$ subject to  the constraints $A \geq 0$,

$B \geq 0$, $C \geq 0$ and $A+B+C=\pi$ can be done if instead of the AM-GM inequality we use a stronger inequality, called Jensen’s inequality. It is stated as follows:

Theorem.

Suppose $h(x)$ is a twice differentiable, real-valued function on an interval $[a,b]$ and that $h^{''}(x)>0$ for all $a. Then, for every positive integer m and for all points $x_{1}, x_{2}, \ldots x_{m}$ in $[a,b]$, we have

$h(\frac{x_{1}+x_{2}+\ldots+x_{m}}{m}) \leq \frac{h(x_{1})+h(x_{2})+h(x_{3})+\ldots+h(x_{m})}{m}$

Moreover, equality holds if and only if $x_{1}=x_{2}=\ldots=x_{m}$. A similar result holds if

$h^{''}(x)<0$ for all $a except that the inequality sign is reversed.

What this means is that the value of assumed by the function h at the arithmetic mean of a given set of points in the interval $[a,b]$ cannot exceed the arithmetic mean of the values assumed by h at these points, More compactly, the value at a mean is at most the mean of values if $h^{''}$ is positive in the open interval $(a,b)$ and the value at a mean is at least the mean of values if $h^{''}$ is negative on it. (Note that $h^{''}$ is allowed to vanish at one or both the end-points of the interval $[a,b]$.)

A special case of Jensen’s inequality is the AM-GM inequality.

Jensen’s inequality can also be used to give easier proofs of certain other trigonometric inequalities whose direct proofs are either difficult or clumsy. For example, applying Jensen’s inequality to the function $h(x)=\sin{x}$ on the interval $[0,\pi]$ one gets the following result. (IITJEE 1997)

If n is a positive integer and $0 for $i=1,2,\ldots, n$, then

$\sin{A_{1}}+\sin{A_{2}}+\ldots+\sin{A_{n}} \leq n \sin{(\frac{A_{1}+A_{2}+\ldots+A_{n}}{n})}$.

More later,

Nalin Pithwa

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