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# Inequality of fractions - TOMATO Subjective 12

## Inequality of fractions

An inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions. It is used most often to compare two numbers on the number line by their size. In this post we are going to discuss a problem on inequality of fractions.

## Try the problem

This problem is from Indian Statistical Institute

Let $$x_n = \frac {1}{2} \cdot \frac{3}{4} \cdot \frac {5}{6} \cdots \frac{2n-1}{2n}$$

Then show that $$x_n \leq \frac {1} {\sqrt{3n+1} }$$

Test of Mathematics at 10 + 2 Level, problem 12, Subjective. This is also an old Math Olympiad Problem and entrance of Indian Statistical Institute (B.Stat, B.Math)

Inequalities (Algebra), Mathematical Induction

6 out of 10

Secrets in Inequalities by Pham Kim Hung

## Use some hints

First Hint

This can be easily proved by mathematical induction.

First, check that the claim is true for n=1

That is $$x_1 = \frac{1}{2} \geq \frac{1}{\sqrt{3\times 1 + 1}} = \frac{1}{2}$$

Now assume that, it is true for n = k. That is, assume $$x_k = \frac {1}{2} \cdots \frac{2k-1}{2k} \leq \frac {1} {\sqrt{3k+1} }$$

Finally, prove that it is true for n = k +1

Second Hint

Let us write the expression for n = k+1.

$$x_{k+1} = \frac {1}{2} \cdots \frac{2k-1}{2k} \cdot \frac{2(k+1)-1}{2(k+1)}$$

Since $$x_k = \frac {1}{2} \cdots \frac{2k-1}{2k} \leq \frac {1} {\sqrt{3k+1} }$$

We can replace the first portion of ( x_{k+1} ) by ( \frac {1} {\sqrt{3k+1} } ) and the resultant quantity will be greater than ( x_{k+1} )

Particularly

$$x_{k+1} \leq \frac {1} {\sqrt{3k+1}} \cdot \frac{2(k+1)-1}{2(k+1)} =\frac {1} {\sqrt{3k+1}} \cdot \frac{2k+1}{2k+2}$$

Third Hint

We want to show that $$x_{k+1} \leq \frac{1} {\sqrt{3\times (k+1) + 1} }$$

We found something even larger than ( x_{k+1} ), that is ( \frac {1} {\sqrt{3k+1} } \cdot \frac{2k+1}{2k+2} )

If we can show that this larger quantity is smaller than ( \frac{1} {\sqrt{3\times (k+1) + 1} } ) then certainly ( x_{k+1} ) will be smaller than ( \frac{1} {\sqrt{3\times (k+1) + 1 }} ).

Final Hint

Suppose otherwise (proof by contradiction).

Assume that ( \frac {1} { \sqrt {3k+1} } \cdot \frac{2k+1}{2k+2}  >  \frac{1} { \sqrt{ 3\times (k+1) + 1 } } )

For example, square both sides and cross multiply to find

$$( \sqrt{3k+4})^2 \times (2k+1)^2 > (\sqrt{3k+1})^2 \times (2k+2)^2$$

Expand and simplify both sides to find $$12 k^3 + 28 k^2 + 19k + 4 > 12 k^3 + 28 k^2 + 20k + 4$$

Cancelling everything we are left out with $$19k > 20k$$

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