Binomials and Inequalities

Binomial expansion [2]

Largest coefficient

B2, 2011

Show that the power of with the largest coefficient in the polynomial is 8. In other words, if we write the given polynomial as then the largest coefficient is .

Solution

The coefficient . Consider the ratio of two consecutive terms: . The ratio up to and strictly less than 1 for . Hence, the sequence of coefficients is bitonic with the peak occurring at .

Coefficient ratio

A7, 2015

(i) By the binomial theorem where are appropriate constants. Write the value of for which is the largest among the 11 terms in this sum.

(ii) For every natural number let where and are integers. Calculate .

Solution

(i) . Consider the ratio:

This ratio is till and from onwards. Similar to problem B2, 2011.

(ii) 32. Using binomial expansion see that where the sign depends on the parity of As since Thus and so

Inequalities [4]

AM-GM inequality

A4, 2011

Given positive real numbers whose product , what can you say about their sum ?

can be any positive number.
.
is unbounded above.
is bounded above.

Solution

is unbounded above.

The first inequality follows from AM-GM inequality. To see why is unbounded, set , and the rest of to 1. The sum for any .

AM-GM inequality II

B5, 2012

Suppose .

(a) For any real show that What kind of a number is ?

Solution

Without loss of generality, we may assume that . Now use the AM-GM inequality:
since , the number must be a non-real complex number.

Combinatorial and calculus inequalities

B2b, 2021

Prove or disprove:
(i)
(ii) for all

Solution

(i) False.

Comment: A similar problem was asked in mock test #6 (Problem B2).

(ii) True.
Let us prove the left hand side inequality.

for

and

for

. This means

is decreasing in the interval

and increasing in the interval

. It reaches the minimum at

which is zero. Hence

for all

.
The right hand side inequality can be proved similarly.

Again,

attains its minima of zero at

Points on a sphere

B2, 2015

Let and be real numbers with

(a) Prove the inequality

(b) Also find the smallest possible value of .

Specify exactly when the smallest and the largest possible value is achieved.

Solution

We have where

Let us examine possible values of in view of the constraint .

We have e.g. because

Adding we get because . Thus . So

Note that equalities are achieved precisely when and .

Thus altogether we have to find extrema of the odd function over the interval The critical points are when i.e. Thus we need to see only and Therefore Moreover, precisely when

In each case, this gives a line segment in the -plane joining and Note that both these segments lie within the circle so each point on them leads to two valid points on the unit sphere.