Test of Mathematics Solution Subjective 157 -Limit of a product

This is a Test of Mathematics Solution Subjective 157 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Evaluate $ \mathbf {\displaystyle \lim_{n \to \infty } \{ (1 + \frac{1}{2n}) (1 + \frac{3}{2n} )(1+ \frac{5}{2n}) ... (1+ \frac{2n-1}{2n})\}^{\frac{1}{2n}} }$

Discussion

Let $ \mathbf { y = \{ (1 + \frac{1}{2n}) (1 + \frac{3}{2n} )(1+ \frac{5}{2n}) ... (1+ \frac{2n-1}{2n})\}^{\frac{1}{2n}} }$

Then $ \mathbf { \log (y) = \frac{1}{2n}\{{ \log (1 + \frac{1}{2n}) + \log (1 + \frac{3}{2n} )+ \log (1+ \frac{5}{2n}) + ... + log (1+ \frac{2n-1}{2n})} }\}$

This implies $ \log (y) = \frac{1}{2n} { \log (1 + \frac{1}{2n}) + \cdots + \log (1 + \frac{2n}{2n}) } $ - $ \frac{1}{2n} { \log (1 + \frac{2}{2n}) + \log (1 + \frac{4}{2n} )+ \log (1+ \frac{6}{2n}) + ... + \log (1+ \frac{2n}{2n}) } $

$ \displaystyle \lim_{n \to \infty} \log (y) = \int_0^1 \log (1+x) dx - \frac{1}{2} \int_0^{1} \log(1+x) dx = \frac{1}{2} \int_{0}^1 \log (1+x) dx = \log 2 - \frac {1}{2} $

Therefore answer will be $e^{\log 2 - \frac {1}{2}}$

Test of Mathematics Solution Subjective 155 -The Lim 1/(n+r) Problem

This is a Test of Mathematics Solution Subjective 155 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Evaluate: $ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n})$

Solution

As the title suggests the modification of this problem will be, that we will solve a more general series and then use a specific value to arrive at the solution of this problem.

First let us consider the following limit:

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+kn})$

Observe carefully that using k=1 in this limit, we get the limit that has been asked to evaluate.

Now

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{1}{n+r})$

$ = \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}})$

$ = \lim_{n\to\infty}\frac{1}{n} (\sum_{r=1}^{kn} \frac{1}{1+\frac{r}{n}})$

Let's substitute $ \frac{r}{n} = x => dr = ndx$

Now we can change the sum to an integral

$ => \lim_{n\to\infty} (\sum_{r=1}^{kn} \frac{\frac{1}{n}}{1+\frac{r}{n}}) = \lim_{n\to\infty} \frac{1}{n}*n\int_{0}^{k} \frac{1}{1+x} dx $

$ = \lim_{n\to\infty}( log |x+1|_{k} - log |x+1|_{0})$

$ = \lim_{n\to\infty} log |k+1|$

$ = log |k+1|$ (As the term is an 'n' free term)

So we see the solution is $ = log |k+1|$

Substituting k=1, we get

$ \lim_{n\to\infty} (\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{n+n}) = \log {2}$

Test of Mathematics Solution Subjective 150 - Maximum of nth roots of n

This is a Test of Mathematics Solution Subjective 150 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Find the maximum among $ \mathbf { 1 , 2^{1/2} , 3^{1/3} , 4^{1/4} , ... }$ .

Solution

Consider the function $ \mathbf { f(x) = x^{1/x} }$ . We employ standard techniques to compute the maxima.

Take logarithm on both sides we have $ \mathbf { \log f(x) = \frac{1}{x} \log x }$ . Next find out the derivative:

$ \mathbf {\frac {1}{f(x)} f'(x) = \frac{-1}{x^2} \log x + \frac{1}{x}\cdot\frac{1}{x} implies f'(x) = f(x) \cdot \frac{1}{x^2} (1 - \log x) }$

Since $ \mathbf { f(x) = x^{1/x} }$ is always positive for positive x and so is $ \mathbf {\frac{1}{x^2}}$ sign of the derivative depends only on (1-logx). Hence the derivative is 0 at x = e (2.71 approximately), positive before that and negative after that. Hence the function has a maxima at x = e.

We check the values at x=2 and x=3 and easy computations show that $ \mathbf { 3^{1/3} > 2^{1/2} }$. Hence $ \mathbf {3^{1/3} }$ is the largest value.

Special Note

One may ask for a non calculus proof of this problem. The basic idea is to understand that the inequality
$ \mathbf { n^{1/n} > (n+1)^{1/n+1}\Rightarrow n^{n+1} > (n+1)^n \Rightarrow n\cdot n^n > (n+1)^n \\ \Rightarrow n > \frac{(n+1)^n}{n^n}\Rightarrow n > (1+ \frac{1}{n})^n }$

It is easy to show that the quantity $ \mathbf { (1+ \frac{1}{n})^n }$ lies within 2 and 3 for all values of n. Hence the inequality $ \mathbf {n > (1+ \frac{1}{n})^n }$ is true for n > 3. The result follows.

Test of Mathematics Solution Subjective 144 - Finding a Function's Upper Bound

This is a Test of Mathematics Solution Subjective 144 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta

Problem

Suppose $ f(x)$ is a real valued differentiable function defined on $ [1,\infty)$ with $ f(1)=1$. Suppose moreover $ f(x)$ satisfies

$ f'(x) = \frac {1}{x^2+f^2(x)}$

Show that $ f(x) \leq 1+\frac{\pi}{4}$ for every $ x \geq 1$

Solution

As the question doesn't require us to find an exact solution rather just an upper bound, we can easily find it by manipulating the given statement after establishing certain properties of $ f(x)$.

We see that $ f'(x)>0$ for all $ x$ which means $ f(x)$ is an increasing function.

As the domain is $ [1,\infty)$ we can say that $ f(x)\geq f(1) $ for all $ x$.

$ => f^2(x)\geq f^2(1)$

$ => x^2+f^2(x)\geq x^2+f^2(1)$ (as $ x^2> 0$)

$ => \frac{1}{x^2+f^2(x)}\leq \frac{1}{x^2+f^2(1)}$

$ => f'(x)\leq \frac{1}{x^2+f^2(1)}$

Integrating both sides from 1 to $ x$

$ => \int_{1}^{x}f'(x)\leq \int_{1}^{x}\frac{1}{x^2+f^2(1)}\leq \int_{1}^{\infty}\frac{1}{x^2+f^2(1)}$

As $ f(1)=1$ we have,

$ => \int_{1}^{x}f'(x)\leq \int_{1}^{\infty}\frac{1}{x^2+1}$

$ => f(x) - f(1) \leq tan^{-1}\infty - tan^{-1}1$

$ => f(x) - f(1) \leq \frac{\pi}{2} - \frac{\pi}{4}$

Substituting $ f(x) = 1$

$ => f(x) \leq 1+\frac{\pi}{4}$

Hence Proved.

Important Resources:

Test of Mathematics Solution Subjective 127 -Graphing relations

This is a Test of Mathematics Solution Subjective 127 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

Find all (x, y) such that sin x + sin y = sin (x+y) and |x| + |y| = 1

Discussion

|x| + |y| =1 is easier to plot. We have to treat the cases separately.

First quadrant: x +y = 1
Second quadrant: -x + y = 1 (since |x| = -x when x is negative)
Third Quadrant: -x-y =1
Fourth Quadrant: x - y =1

Now we work on sin x + sin y = sin (x + y).

This implies $ \displaystyle{2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x-y} {2} \right ) = 2 \sin \left ( \frac{x+y} {2} \right ) \cos \left ( \frac{x+y} {2} \right ) }$. Hence we have two possibilities:

$ \displaystyle{\sin \left ( \frac{x+y} {2} \right ) = 0 }$ OR
$ \displaystyle{\cos \left ( \frac{x+y} {2} \right ) - \cos \left ( \frac{x-y} {2} \right ) = 0 }$ or $ \displaystyle { \sin \frac{x}{2} \sin \frac{y}{2} } = 0 $

The above situations can happen when when

$ \displaystyle{ \frac{x +y}{2} = k \pi } $ or $ \displaystyle{\frac {x}{2} = k \pi }$ or $ \displaystyle{ \frac{y}{2} = k \pi }$, where k is any integer.

Thus we need to plot the class of lines $ \displaystyle{ x + y = 2 k \pi } $, $ \displaystyle{ x = 2k\pi } $ and $ \displaystyle{ y = 2k\pi } $, and consider the intersection points of these lines with the graph of |x| + |y| = 1.

Clearly only for k=0, such intersection points can be found.

Hence required points are (0,1), (0,-1), (1,0), (-1,0), (1/2, -1/2), (-1/2, 1/2).

Chatuspathi

What is this topic: Graphing
What are some of the associated concept: Trigonometric Identities
Where can learn these topics: I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Calculus’ module.
Book Suggestions: Play With Graphs (Arihant Publication)

Test of Mathematics Solution Subjective 126 - Graphs of Absolute Value Functions

This is a Test of Mathematics Solution Subjective 126 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

Sketch, on plain paper, the regions represented, on the plane by the following:

(i) $ |y| = \sin x $

(ii) $ |x| - |y| \ge 1 $

Solution

First we need to understand what |y| signifies. It is the absolute value of y, that is it is +y when y is positive and -y when y is negative.

Lets test with $ x = \frac{\pi}{6} $. Clearly then sin x = 1/2. This implies $ |y| = 1/2 $ or $ y = 1/2, -1/2 $.

Again let us test with $ x = \pi + \frac{\pi}{6} $. Then $ \sin (\pi + \frac{\pi}{6}) = - \frac{1}{2} $ implying $ |y| = - \frac{1}{2} $. But this is impossible as absolute value cannot be negative.

Using these observations we get a clear idea about what is happening.

The values of x where sin (x) is positive (that is when $ \displaystyle {\frac {4k \pi}{2} \le x \le \frac {(4k+2) \pi }{2} }$, where k is any integer), we draw the graph of y = sin x and reflect it about x axis (as y = sin x and y = - sin x both satisfies the equation).
The values of x where sin (x) is negative (that is when $ \displaystyle {\frac {(4k+2) \pi}{2} \le x \le \frac {(4k+4) \pi }{2} }$, where k is any integer), the given relation is not defined as absolute value of y cannot be negative.

For Part (ii)

Notice that |x| gives distance of a point from y axis and |y| gives distance of a point from x axis.

Let us split the problem into cases:

x, y both non negative (first quadrant). Then |x| = x, |y| = y, implying $ |x| - |y| \ge 1 $ is same as $ x - y \ge 1 $
$ x \le 0, y \ge 0 $ implies |x| = -x. Thus in second quadrant the inequality becomes $ -x -y \ge 1$ or $ x+y \le -1 $
$ x \le 0, y \le 0 $ implies |x| = -x, |y| = -y. Thus in third quadrant the inequality becomes $ -x +y \ge 1$ or $ x-y \le -1 $
$ x \ge 0, y \le 0 $ implies |x| = x, |y| = -y. Thus in fourth quadrant the inequality becomes $ x + y \ge 1$

Chatuspathi:

What is this topic: Graphing of functions, inequalities
What are some of the associated concept: Absolute value functions, inequalities
Where can learn these topics: I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Calculus’ module.
Book Suggestions: Play with Graphs (Arihant Publication)

Test of Mathematics Solution Subjective 125 - Function on Natural Numbers

This is a Test of Mathematics Solution Subjective 125 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

Let $ f: \mathcal{N} to \mathcal{N} $ be the function defined by f(0) = 0, f(1) = 1, and f(n) = f(n-1) + f(n-2), for $ n \ge 2 $, where $ \mathcal {N} $ is the set of all non negative integers. Prove the following results:

f(n) < f(n+1) for all $ n \ge 2 $
There exists precisely four non-negative integers n for which f(f(n)) = f(n).
f(5n) is divisible by 5, for all n.

Solution

This problem gives us an opportunity to discuss Fibonacci Sequence (which is defined by this function) and solving linear recursions. However none of these two discussions are necessary to solve the problem. Hence we leave those discussions for class and focus on the solution itself:

Clearly f(2) = 1, and f(3) = f(2) + f(1) = 2. Hence f(3) is positive. We use strong form of induction and assume that up to n, all f(n) is positive.

Then f(n+1) = f(n) + f(n-1) .

Since f(n-1) is positive. Hence f(n+1) > f(n). This solves first part.

For second part we observe

$f(f(0)) = f(0)$
$f(f(1)) = f(1) (as f(1) = 1)$
As $f(2) = f(0) + f(1) = 1$, we have $f(f(2)) = f(1) = 1 = f(2)$.
Next observe that
- $f(3) = f(2) + f(1) = 1 + 1= 2$
- $f(4) = f(3) + f(2) = 2 + 1 = 3$
- $f(5) = f(4) + f(3) = 3 + 2 = 5$
- So $f(f(5)) = f(5)$ as $f(5) = 5$

So we got four numbers 0, 1, 2, 5, for which $f(f(n)) = f(n)$ by observation. We wish to show that this does not happen for any value of n greater than 5.

Clearly, if $f(f(n)) = f(n)$, then $f(n) = n$. We will show that $f(n) > n$ for all n greater than 5. (If we can show that then it will automatically show that f(n) is not equal n for any value greater than 5, implying f(f(n)) not equal to f(n) for any other value apart from the ones that we have found earlier by trial and error).

Note that $f(6) = f(5) + f(4) = 5 + 3 = 8$. That is $f(6) > 6$. Similarly $f(7) = f(6)+ f(5) = 8+5 = 13$ implying $f(7)> 7$.

We will again use strong for of induction. Suppose $f(k) > k$ for all values of k from 6 to n $ n \ge 7 $. Using this assumption we will show that $f(n+1) > n+1$.

$f(n+1) = f(n) + f(n-1)$. By induction $f(n) > n$ and $f(n-1) > n-1$. This implies $f(n+1) > n + n -1 = 2n -1$.

But $2n-1 > n+1$ (as n > 2). Hence it follows that $f(n+1) > n+1$. This solves second part.

Finally we wish to show that 5 divides f(5n).

We again use induction. First note that f(5) = 5 hence divisible by 5. Assume that the claim is true for n-1. That is 5 divides f(5(n-1)). Using this assumption we wish to show that 5 divides f(5n). Now note that:

$f(5n)$

= $f(5n-1) + f(5n-2)$

= $f(5n-2) + f(5n-3) + f(5n-3) + f(5n-4)$

= $f(5n-3) + f(5n-4) + f(5n- 4) + f(5n-5) + f(5n-4) + f(5n- 5) + f(5n - 4)$

= $f(5n-4) + f(5n-5) + f(5n-4) + f(5n- 4) + f(5n-5) + f(5n-4) + f(5n- 5) + f(5n - 4)$

= $5 f(5n-4) + 3 f(5n-5)$

Clearly 5 $f(5n-4)$ is divisible by 5. Also $3 f(5n-5)$ is divisible by 5 as by inductive assumption $f(5n-5)$ is divisible by 5. Hence their sum 5 $f(5n-4) + 3 f(5n-5)$ is divisible by 5. This shows 5 divides 5 divides $f(5n)$.

This completes the proof of third part.

Comments:

Actually the problem is more fun if we bring in ideas of fibonacci numbers and work with it's analytical form by solving the linear recurrence given here.

Chatuspathi:

What is this topic: Mathematical Induction
What are some of the associated concept: Fibonacci Numbers, Linear Recurrences
Where can learn these topics: I.S.I. & C.M.I. Entrance course of Cheenta, discusses these topics in the ‘Number Theory’ module.
Book Suggestions: Elementary Number Theory by David Burton

Test of Mathematics Solution Subjective 124 - Graph sketching

This is a Test of Mathematics Solution Subjective 124 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

Sketch on plain paper, the graph of $ y = \frac {x^2 + 1} {x^2 - 1} $

Solution

There are several steps to find graph of a function. We will use calculus to analyze the function. Here y=f(x)

Domain: The function is defined at all real numbers except x =1 and x = -1 which makes the denominator 0.
Even/Odd: Clearly f(x) = f(-x). Hence it is sufficient to investigate the function for positive values of x and then reflect it about y axis.
Critical Points: Next we investigate the critical points. Critical Points are those values of x for which the first derivative of f(x) is either 0 or undefined. Since $ \displaystyle{y = \frac {x^2 + 1} {x^2 - 1}}$, then $ \displaystyle{f'(x) = \frac {\left(\frac{d}{dx}(x^2 + 1)\right)(x^2 -1) - \left(\frac{d}{dx}(x^2 - 1)\right)(x^2 + 1)} {(x^2 - 1)^2} }$.
This implies $ \displaystyle{f'(x) = \frac{2x^3 - 2x - 2x^3 - 2x}{(x^2 - 1)^2 } = -\frac{4x}{(x^2 - 1)^2 }}$
Hence critical points are x =0 , 1, -1
Monotonicity: The first derivative is negative for all positive values of x (note that we are only investigating for positive x values, since we can then reflect the picture about y axis as previously found). Hence the function is 'decreasing' for all positive value of x.
Second Derivative: We compute the second derivative to understand a couple things:
1. convexity/concavity of the function
2. examine whether the critical points are maxima, minima, inflection points.
  $ \displaystyle{f''(x) = \frac {4(3x^2 +1)}{(x^2-1)^3}}$
  Clearly $ f''(0) = -4 $ implying at x = 0 we have local maxima. Since f(0) = - 1, we have (0, -1) as a local maxima.
  Also the second derivative is negative from x = 0 to x = 1 and positive after x = 1. Hence the curve is under-tangent (concave) from x = 0 to x = 1, and above-tangent (convex) from x =1 onward.
Vertical Asymptote: We next examine what happens near x = 1. We want to know what happens when we approach x=1 from left and from right. To that end we compute the following limits:
1. $ \displaystyle {\lim_{x to 1^{-} } \frac{x^2 +1}{x^2-1} = -\infty}$ (since the denominator gets infinitesimally small with a negative sign, and numerator is about 2)
2. $ \displaystyle {\lim_{x to 1^{+} } \frac{x^2 +1}{x^2-1} = +\infty}$ (since the denominator gets infinitesimally small with a positive sign, and numerator is about 2)
Horizontal Asymptote: Finally we examine what happens when x approaches $ + \infty $. To that end we compute the following:
$ \displaystyle { \lim_{x to + \infty} \frac {x^2 +1}{x^2-1}= \lim_{\frac{1}{x} to 0} \frac {1+ \frac{1}{x^2}}{1-\frac{1}{x^2}} = 1} $
Drawing the graph:
1. Local Maxima at (0,-1)
2. Even function hence we draw for positive x values and reflect about y axis
3. Vertical asymptote at x =1
4. From x = 0 to 1, the function decreasing to negative infinity, staying under tangent all the time.
5. From x = 1 to positive infinity, the function decreases from positive infinity to 1 staying above tangent all the time.
6. Horizontal Asymptote at y= 1

Chatuspathi:

What is this topic: Graph Sketching using Calculus
What are some of the associated concept: Maxima, Minima, Derivative, Convexity, Concavity, Asymptotes
Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the 'Calculus' module.
Book Suggestions: Calculus of One Variable by I.A. Maron, Play with Graph (Arihant Publication)

Graphing integer value function | Tomato Subjective 117

This is a subjective problem from TOMATO based on Graphing integer value function.

Problem: Graphing integer value function

Let [x] denote the largest integer (positive, negative or zero) less than or equal to x. Let $y= f(x) = [x] + \sqrt{x - [x]} ,s=2 $ be defined for all real numbers x.

(i) Sketch on plain paper, the graph of the function f(x) in the range $-5 \le x \le 5 ,s=2$
(ii) Show that, any given real number $y_0 ,s=2 $, there is a real number $x_0 ,s=2 $ such that $y_0 = f(x_0) ,s=2 $

Discussion:

First note that $\sqrt{x - [x]} ,s=2 $ is same as $\sqrt{t} , 0\le t \le 1 ,s=2 $.

It's graph between 0 to 1

Clearly [x] part only increments (or decrements) it by integer quantity as [x] is constant between any two integers. That for any integer k for all ( x \in (k, k+1) ).
$f(x) = k +\sqrt{t} ,s=2 $ , $t\in(0,1) ,s=2 $.

Finally consider and arbitrary value $ y_0 ,s=2 $. We take $x_0 = [y_0] + (y - [y_0])^2 ,s=2$. Then $f(x_0) = [x_0] + \sqrt(x - [x_0] = [y_0] + \sqrt{(y - [y_0])^2} = y_0 ,s=2 $ (since $0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1 ,s=2 $ )

Chatuspathi:

What is this topic: Graphing of functions
What are some of the associated concept: Greatest Integer Function
Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Calculus’ module.
Book Suggestions: Play with Graphs, Arihant Publication

Test of Mathematics Solution Subjective 116 - Angles in a Triangle

This is a Test of Mathematics Solution Subjective 116 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Problem

If A, B, C are the angles of a triangle, then show that $ \displaystyle { \sin A + \sin B - \cos C \le \frac {3 \sqrt{3}}{2}}$

Solution:

$ \displaystyle { \sin A + \sin B - \cos C } $
$ \displaystyle { = \sin A + \sin B - \cos (\pi - (A+B)) } $
$ \displaystyle { = \sin A + \sin B + \sin (A+B) } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} \cos \frac{(A-B)}{2} + 2\sin \frac{(A+B)}{2} \cos \frac{(A+B)}{2} } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} \left( \cos \frac{(A-B)}{2} + \cos \frac{(A+B)}{2}\right ) } $
$ \displaystyle { = 2\sin \frac{(A+B)}{2} 2\cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\sin \frac{(\pi -C)}{2} \cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\sin \left(\frac{\pi}{2} - \frac{C}{2} \right) \cos \frac{A}{2} \cos \frac{B}{2} } $
$ \displaystyle { = 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} } $

We apply Jensen's Inequality and Arithmetic Mean - Geometric Mean inequality here. Since cosine function is concave in the interval $ [0, \frac{\pi}{2} ] $, we have
$ \displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\cos \frac{C}{2} + \cos \frac{A}{2} +\cos \frac{B}{2}}{3} \le \cos \left ( \frac{1}{3}\times \frac{A}{2} + \frac{1}{3}\times \frac{B}{2} + \frac{1}{3}\times \frac{C}{2} \right ) } $
This implies $ \displaystyle { \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \left ( \frac{A+B+C}{6}\right ) } $
$ \displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \cos \frac{\pi}{6} } $
$ \displaystyle { \Rightarrow \left (\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \right )^{\frac{1}{3}} \le \frac{\sqrt{3}}{2} } $
$ \displaystyle { \Rightarrow \cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{8} } $
$ \displaystyle { \Rightarrow 4\cos \frac{C}{2} \cos \frac{A}{2} \cos \frac{B}{2} \le \frac{3\sqrt{3}}{2} } $

Chatuspathi:

What is this topic: Properties of triangle
What are some of the associated concept: Jensen Inequality, Arithmetic Mean-Geometric Mean Inequality
Where can learn these topics: I.S.I. & C.M.I. Entrance Course of Cheenta, discusses these topics in the ‘Inequality’ module.
Book Suggestions: 'Secrets in Inequality' by Pam Kim Hung