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The image is a front cover of a book named Real & Complex Analysis by Walter Rudin. Here you can find Jensen’s Inequality also this book is very useful for the preparation of NBHM Entrance.

Also Visit: College Mathematics Program of Cheenta

## Problem:

Let (n\in\mathbb{N}), (n\ge 2). Let (x_1,x_2,…,x_n\in(0,\pi)). Set (x=\frac{x_1+x_2+…+x_n}{n}). Which of the following are true?

A) (\prod_{k=1}^{n} sinx_k \ge sin^nx )

B) (\prod_{k=1}^{n} sinx_k \le sin^nx )

C) Neither (A) or (B) is necessarily true.

## Discussion:

Well, the title of this post gives it away. You have to use Jensen’s inequality. But before that, notice that the first inequality is easily seen to be a false one. For I can have the midpoint of two points in ((0,\pi)) as (\pi/2) which will mean the right-hand side is one and the left-hand side is less than 1.

There are many different forms of what is called Jensen’s inequality, in measure theory context the inequality is stated in terms of integrals. But we will use the sum form as stated below.

Let (a_1,…,a_n) be non-negative real numbers such that (a_1+a_2+…+a_n=1). Suppose that (f) is a convex function. Then

$$f(a_1x_1+a_2x_2+…+a_nx_n) \le a_1f(x_1)+a_2f(x_2)+…+a_nf(x_n)$$

where (x_i  )s are points in the domain of (f).

Note that if I apply this to two points (x_1,x_2) then I will just get the definition of a convex function. This can serve as a check if confused about the direction of the inequality.

Coming back to our problem, observe that we can not directly apply this inequality to (sin) function on ((0,\pi)). Because (sin) is not convex but concave in ((0,\pi)). There are many ways to verify the concavity of (sin) in this part, the easiest one being drawing the graph. Alternately, (sin”(x)=-sin(x)<0) on ((0,\pi)), meaning that the derivative of (sin) is a decreasing function, meaning (sin) is concave in this region.

Okay, once we have a concave function (g) ((g(x)=sinx)), we have a convex function also. Namely (-g). Once again, the easiest way to see this is to draw the graph. Alternatively, we have (g((1-a)x_1+ax_2) \ge (1-a)g(x_1)+ag(x_2) ) for (a\in [0,1]). So (-g((1-a)x_1+ax_2) \le (1-a)(-g)(x_1)+a(-g)(x_2) ) which proves the convexity of (-g).

So, to sum up, we can apply Jensen’s inequality to (-sin) on ((0,\pi)).

This gives

$$sin(\frac{x_1}{n}+\frac{x_2}{n}+…+\frac{x_n}{n}) \ge \frac{sinx_1}{n}+\frac{sinx_2}{n}+…+\frac{sinx_n}{n}$$

Here we have used each (a_i=1/n) ,(i=1,…,n).

Now notice that the right-hand side has each term non-negative. So we are allowed to apply AM-GM inequality. This gives

$$\frac{sin(x_1)+…+sin(x_n)}{n} \ge ( \prod_{k=1}^{n} sinx_k )^{1/n}$$

Combining the last two inequalities we get the inequality

(\prod_{k=1}^{n} sinx_k \le sin^nx ).