Inequality Problem | ISI - MSQMS 2018 | Part B | Problem 4

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality and Combinatorics.

INEQUALITY | ISI 2018 | MSQMS | PART B | PROBLEM 4

Show that $\sqrt{C_{1}}+\sqrt{C_{2}}+\sqrt{C_{3}}+\ldots+\sqrt{C_{n}} \leq 2^{n-1}+\frac{n-1}{2}$ where
$C_k={n\choose k}$

Key Concepts

INEQUALITIES

COMBINATORICS

Try with Hints

Use Cauchy Schwarz Inequality $\left(\displaystyle\sum_{i} a_{i} b_{i}\right)^{2} \leq\left(\displaystyle\sum_{i} a_{i}^{2}\right)\left(\displaystyle\sum_{i} b_{i}^{2}\right)$

Apply Cauchy Schwarz Inquality in two sets of real numbers ($\sqrt C_1$,$\sqrt C_2$,.....,$\sqrt C_n$)and ($1$,$1$,$1$,......$1$)

($C_1+C_2+$........$+C_n$)($1+1+$......$+1$) $\geq $ ($\sqrt C_1+\sqrt C_2+.........+\sqrt C_n$)

($2^n-1$)$n \geq $ ($\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n$)$^2$

$\sqrt C_1+\sqrt C_2+$..........$+\sqrt C_n \leq \sqrt n\sqrt (2^n-1)$

The proof is still not done,why don't you try the remaining part yourself?

We know AM $\geq$ GM

i.e

For $n$ positive quantities $a_{1}, a_{2}, \dots, a_{n}$
$$
\frac{a_{1}+a_{2}+\ldots+a_{n}}{n} \geq \sqrt[n]{a_{1} a_{2} \cdot \cdot a_{n}}
$$
with equality if and only if $a_{1}=a_{2}=\ldots=a_{n}$

Now you have all the ingredients,why don't you cook it yourself? I firmly believe that you can cook a food tastier than mine.

$\frac{n+2^n-1}{2} \geq \sqrt n\sqrt {2^n-1}$

Thus,$\sqrt C_1+\sqrt C_2+$........$+\sqrt C_n \leq \frac {n+2^n-1}{2}$

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TIFR 2014 problem 6 Solution - Inequality in Integration


TIFR 2014 Problem 6 Solution is a part of TIFR entrance preparation series offered by Cheenta. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects. Generally, the exams scheduled in December.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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Problem

Let (f:[0,1] \to \mathbb{R} ) be a continuous function. Which of the following statement is always true?

A. ( \int_{0}^{1} f^2(x)dx = (\int_{0}^{1}f(x)dx)^2 )

B.  ( \int_{0}^{1} f^2(x)dx \le (\int_{0}^{1}|f(x)|dx)^2 )

C.  ( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 )

D.  ( \int_{0}^{1} f^2(x)dx < (\int_{0}^{1}f(x)dx)^2 )

Discussion:

We first recall the Cauchy-Schwartz inequality for an inner product space (V) and two vectors (a,b\in V)

( <a,b> \le ||a||||b|| ).

Here, we also remember the fact that (C[0,1]) (the set of all continuous real (/complex) valued functions on [0,1] ) forms an inner product space with respect to the inner product

( <f,g>=  \int_{0}^{1} f(x)g(x)dx ) (We are only taking real valued functions so we neglect the conjugation...)

We want to apply this inequality to suitable functions so that we get some inequality from the options above.

Let's try (|f|) and (1) (the constant function).

We get: ( \int_{0}^{1} |f(x)|1dx \le (\int_{0}^{1} |f(x)|^2dx)^{1/2} (\int_{0}^{1} 1^2dx)^{1/2} )

Squaring, we get

( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 ).

So option C ( \int_{0}^{1} f^2(x)dx \ge (\int_{0}^{1}|f(x)|dx)^2 ) is correct.

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Some Direct Inequalities | TOMATO Subjective 80

This is a beautiful problem based on Some Direct Inequalities from Test of Mathematics Subjective Problem no. 80.

Problem: Some Direct Inequalities

If \(a,b,c\) are positive numbers, then show that

\(\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}+\frac{a^2+b^2}{a+b}\geq a+b+c\)

Solution: This problem can be solved using a direct application of the Titu's Lemma but we will instead prove the lemma first using the Cauchy-Schwarz inequality.

According to the Cauchy-Schwarz inequality we have,

\(\left(a_1^2+a_2^2+\cdots+a_n^2\right)\left(b_1^2+b_2^2+\cdots+b_n^2\right)\ge \left(a_1b_1+a_2b_2+\cdots+a_nb_n\right)^2 \)

Replacing \(a_i\to \dfrac{a_i}{\sqrt{b_i}}\) and \(b_i\to \sqrt{b_i}\) we get,

\(\left(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots +\dfrac{a_n^2}{b_n}\right)\left(b_1+b_2+\cdots+b_n\right)\ge \left(a_1+a_2+\cdots+a_n\right)^2,\)

which is equivalent to

\(\dfrac{a_1^2}{b_1}+\dfrac{a_2^2}{b_2}+\cdots+\dfrac{a_n^2}{b_n}\geq \dfrac{\left(a_1+a_2+\cdots+a_n\right)^2}{b_1+b_2+\cdots+b_n}\)

Now this inequality is referred to as the Titu's Lemma.

This brings us to the problem which can be observed to be a simple application of the lemma. We just need to make the following substitutions.

\(a_1=b, a_2=c\) and \(b_1=b_2=b+c\)

Then we have,

\(\dfrac{b^2}{b+c}+\dfrac{c^2}{b+c}\geq \dfrac{(b+c)^2}{2(b+c)}\)

\(=>\dfrac{b^2+c^2}{b+c}\geq \dfrac{b+c}{2}\)

Thus similarly we have,

\(=>\dfrac{c^2+a^2}{c+a}\geq \dfrac{c+a}{2}\) and \(=>\dfrac{a^2+b^2}{a+b}\geq \dfrac{a+b}{2}\)

Adding the three inequalities we get,

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{b+c}{2}+\dfrac{c+a}{2}+\dfrac{a+b}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq \dfrac{2(a+b+c)}{2}\)

\(=>\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}+\dfrac{a^2+b^2}{a+b}\geq {a+b+c}\)

Hence Proved.

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