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 $$ x_n \leq \frac {1} {\sqrt{3n+1} } $$
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
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
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
$$ x_{k+1} \leq \frac {1} {\sqrt{3k+1}} \cdot \frac{2(k+1)-1}{2(k+1)} =\frac {1} {\sqrt{3k+1}} \cdot \frac{2k+1}{2k+2}$$
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 }} ).
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.
Taught by Olympians, I.S.I. students, alumni, and researchers from leading universities in the world.
Group + One-on-One Classes + 24/7 Doubt Clearing Forum
Publishing original scientific research in a peer-reviewed journal is an achievement many scholars work toward for years. Achieving this feat early in one's academic journey requires exceptional intellectual rigor, creativity, and persistence. Aaditya Punatar and Muskan Sadana have reached this remarkable milestone, successfully publishing their collaborative research paper in the prestigious Journal of Emerging Investigators […]
Her project on quantum-accelerated Steiner trees for 15-minute cities earned a place among the finalists at the world's largest pre-college STEM competition. Prishaa Shrimali of Marymount School of New York, was selected as a Grand Finalist at the 2026 Regeneron International Science and Engineering Fair (ISEF) in Phoenix, Arizona. Her project was Optimizing Urban Accessibility: […]
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's M.Stat Entrance. They ranked within the first 50 in the entire country in these entrances. I.S.I. M.Stat Entrance
In 2026, the following Cheenta students have been successful for Indian Statistical Institute's B.Stat Entrance and Chennai Mathematical Institute's B.Sc. Math Entrance. They ranked within the first 200 in the entire country in these entrances. Most of these students attended the problem solving workshops regularly, which happen 5 days every week. CMI B.Sc. Math Entrance […]