### 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

### Trigonometric Optimization continued

Prove that in any acute angled triangle ABC, $\tan {A}+\tan{B}+\tan{C} \geq 3\sqrt{3}$ with equality holding if and only if the triangle is equilateral. (IITJEE 1998)

Proof:

Suggestion: Try this without reading further! It looks complicated, but need not be so!! Then, after you have attempted whole-heartedly, compare your solution with the one below.

The solution to the above problem is based on the well-known identity:

$\tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}$. For brevity, denote $\tan{A}, \tan{B}, \tan{C}$

by x, y and z respectively. As ABC is acute-angled, x, y, z are all positive and the AM-GM inequality which says

$x+y+z \geq 3{(xyz)}^{1/3}$ can be applied. Taking cubes of both the sides and cancelling

$x+y+z$, (which is positive) this gives $(x+y+z)^{2} \geq 27$. Taking square root we get the desired inequality. If equality is to hold, then it must also hold in the AM-GM inequality, which can happen if and only if

$x=y=z$, that is, if and only if the triangle is equilateral.

Still, this approach requires some caution. Actually, there are so many trigonometric identities that there is no unanimity as to which ones among them are standard enough to be assumed without proof !! But, of course, the IITJEE Examinations, both Mains and Advanced are multiple choice only.

More later,

Nalin Pithwa