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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.

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

Source

Competency

Difficulty level

Suggested book

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

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 } } )

Simplifying this inequality leads to contradiction.

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 $$ 19k > 20k $$

Hence contradiction.

- https://www.cheenta.com/inequality-in-rmo-2019-problem-3-solution/
- https://www.youtube.com/watch?v=mD-FhiUIADo

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