Week -9 | Inequalities – 3

Q-1) Define $S=a_1+a_2+…a_n$ with $a_i \in \mathbb{R}^+$ for $1 \le i\le n$. Show that

for all positive integers $n \ge 2$

Solution:

Using the AM-HM Inequality we have

The RHS on simplification becomes $\frac{n^2}{n-1}$. This is known as the famous Nesbitt’s Inequality

 

Q-2) Find the maximum value of $(7-x)^4(2+x)^5$ with $-2\le x \le 7$.

Solution: Set $p=7-x$ and $q=2+x$. We thus have $p+q=9$, and we need the maximum value of $p^4q^5$.

Applying the AM-GM we have

This gives us

 

Q-3) If $2x^3+ax^2+bx+4=0$ ($a,b$ are positive reals) has 3 real roots, then show that $a+b \ge 6(2^{\frac{1}{3}}+4^{\frac{1}{3}})$

Solution :

Let $\alpha, \beta, \gamma$ be the 3 real roots of the equation.

Gievn that all the coefficients are positive, we know that all the roots are thus negative (Descartes!). So let’s have the substitution $\alpha \rightarrow -\alpha, \beta \rightarrow -\beta, \gamma \rightarrow -\gamma$ with all $\alpha, \beta, \gamma$ positive.

We have $\sum \alpha = \frac{a}{2}$, $\sum \alpha\beta = \frac{b}{2}$ and $\sum \alpha\beta\gamma = 2$. AM-GM gives

$\frac{\alpha + \beta + \gamma}{3} \ge (\alpha \beta \gamma)^{\frac{1}{3}}\Rightarrow \frac{a^3}{216} \ge 2\Rightarrow a\ge 6\times 2^{\frac{1}{3}}$.

Similarly AM-GM on $\sum \alpha\beta$ gives $\frac{b^3}{216}\ge 4\Rightarrow b \ge 6\times 4^{\frac{1}{3}}$. Adding these 2 results we have our proof.

 

Q-4) Show that $(n!)^{\frac{1}{n}} < \frac{n+1}{2}$

Solution :

Consider AM-GM on the first $n$ numbers. We have

$\frac{1+2+3+4…+n}{n}>(1\times 2\times 3…\times n)^{\frac{1}{n}}\Rightarrow \frac{n+1}{2} >(n!)^{\frac{1}{n}}$..

 

Q-5) If $a,b,c,d$ are positive reals such that $a+b+c+d=2$ and $M=(a+b)(c+d)$, then show that $M \le 1$

Solution:

Assuming that $M \ge 1$, we have $a+b \ge \frac{1}{c+d}\Rightarrow a+b+c+d \ge c+d +\frac{1}{c+d}\Rightarrow 2 \ge c+d+\frac{1}{c+d}$, which is a contradiction.

 

Q-6) Let the equation $x^2-3x+p=0$ have 2 positive roots $a,b$. Find the minimum value of $\left ( \frac{4}{a}+\frac{1}{b}\right)$.

Solution:

We have $a+b=3$, applying AM-HM on the numbers $\frac{a}{2}, \frac{a}{2}, b$ we have

 

Q-7) For positive reals $a,b,c$ with $abc=1$ show that $a^{b+c}b^{c+a}c^{a+b} \le 1$

Solution:

Withput loss of generality we can assume $a \le b \le c$, we can make this assumption since the expression $a^{b+c}b^{c+a}c^{a+b}$ is symmetric in $a,b,c$.

We thus have $a \le 1 \le c$. Also $c-b \ge 0$ & $b-a \ge 0$.

This gives us

The required expression we have is :

 

Q-8) If $x,y\in\mathbb{R}^+$, with $m,n \in \mathbb{N}$, show that $\frac{x^ny^m}{(1+x^{2n})(1+y^{2m})} \le \frac{1}{4}$

Solution :

We have

and

Multiplying the 2 we have

An alternate approach may have been by applying the well-known inequality $x+\frac{1}{x} \ge 2$ for $x=x^n$ & $x=y^m$.

 

Q-9) If $a\in \mathbb{R}$ is a root of the equation $x^5-x^3+x-2=0$, show that $\left \lfloor a^6 \right \rfloor =3$. (You may need to read up on the Floor function)

Solution:

Note that from the given equation we have $a[(a^2-1)^2+a^2]=2\Rightarrow a>0$.

We have from AM-GM $a^3+2=a^5+a>2a^3\Rightarrow a^6<4$

Which gives $\left \lfloor a^6 \right \rfloor =3$.

 

Q-10) Let $x>0,y>0$ be reals satisfying $x+y=2$. Show that
a) $x^2y^2(x^2+y^2) \le 2$
b) $x^3y^3(x^3+y^3) \le 2$

Solution:

a) Put $x=1-t$ and $y=1+t$.

We thus have

$x^2y^2(x^2+y^2)=2(1-t^2)(1-t^4) \le 2$.

b) Using $x^3+y^3=(x+y)(x^2-xy+y^2)=2(4-3xy)$, we have on applying of AM-GM on $(4-3xy),xy,xy,xy$

Thus we have

 

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