## Problem 7.2.6

If $a,b,c$ are positive numbers

\begin{align*} (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq 9a^2b^2c^2 \end{align*}

### Solution

We have from $AM \geq GM$,

\begin{align*} \frac{a^2b+b^2c+c^2a}{3} &\geq abc \\ \frac{ab^2+bc^2+ca^2}{3} &\geq abc \end{align*}

Multiplying the above, we get

\begin{align*} (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq 9a^2b^2c^2 \end{align*}

## Problem 7.2.7

Suppose $a_1,\dots,a_n$ are positive numbers and $b_1,\dots,b_n$ is a rearrangement of $a_1,\dots,a_n$, Show that

\begin{align*} \frac{a_1}{b_1} + \cdots + \frac{a_n}{b_n} \geq n \end{align*}

### Solution

We have from $AM \geq GM$,

\begin{align*} \frac{\frac{a_1}{b_1} + \cdots + \frac{a_n}{b_n}}{n} \geq \sqrt[n]{\frac{a_1}{b_1} \cdots \frac{a_n}{b_n}} \end{align*}

As $b_1,\dots,b_n$ is a rearrangement of $a_1,\dots,a_n$, the product on the right hand side above is $1$. Therefore,

\begin{align*} \frac{a_1}{b_1} + \cdots + \frac{a_n}{b_n} \geq n \end{align*}

## Problem 7.2.9

For each integer $n>2$, prove that

a. $\prod_{k=0}^n {n \choose k} < \left( \frac{2^n-2}{n-1} \right)^{n-1}$

b. $n! < \left( \frac{n+1}{2} \right)^n$

c. $1x3x5x \cdots x(2n-1)<n^n$

### Solution

Using $AM>GM$

a. We have

\begin{align*} \frac{\sum_{k=1}^{n-1} {n \choose k}}{n-1} &> \left(\prod_{k=0}^n {n \choose k}\right)^{\frac{1}{n-1}} \\ \implies \left( \frac{2^n-2}{n-1} \right)^{n-1} &> \prod_{k=0}^n {n \choose k} \end{align*}

b. We have

\begin{align*} \frac{\sum_{k=1}^{n}k}{n} &> (n!)^{\frac{1}{n}} \\ \implies \left( \frac{n+1}{2} \right)^n &> n! \end{align*}

c. We have

\begin{align*} \frac{\sum_{k=1}^{n}2k-1}{n} &> (1x3x5x \cdots x(2n-1))^{\frac{1}{n}} \\ \implies (\frac{n^2}{n})^n = n^n &> 1x3x5x \cdots x(2n-1) \end{align*}

## Problem 7.2.10

Given that all roots of $x^6-6x^5+ax^4+bx^3+cx^2+dx+1=0$ are positive, find $a,b,c,d$.

### Solution

Let $u,v,w,x,y,z$ be the roots of the above equation. From $AM\geq GM$, we have

\begin{align*} \frac{u+v+w+x+y+z}{6} = \frac{6}{6} &\geq (uvwxyz)^{1/6} = 1\\ \end{align*}

As we have equality only when $u=v=w=x=y=z=1$, we have

\begin{align*} x^6-6x^5+ax^4+bx^3+cx^2+dx+1 \ \equiv (x-1)^6 = x^6-6x^5+15x^4-20x^3+15x^2-6x+1 \end{align*}

Therefore $a=15,b=-20,c=15,d=-6$.

## Problem 7.3.6

Use the Cauchy-Schwarz inequality to prove that if $a_1,a_2,\dots,a_n$ are real numbers such that $a_1+a_2+\dots+a_n=1$, then $a_1^2+a_2^2+\dots+a_n^2 \geq 1/n$.

### Solution

We have

\begin{align*} (a_1\cdot1+\dots+a_n\cdot1)^2 &\leq (a_1^2 + \dots + a_n^2)(1^2+ \dots+ 1^2) \\ \implies 1 &\leq (a_1^2 + \dots + a_n^2)n \\ \implies \frac{1}{n} &\leq (a_1^2 + \dots + a_n^2) \end{align*}

## Problem 7.3.7

Use the Cauchy-Schwarz inequality to prove the following

a. if $p_1,p_2,\dots,p_n,x_1,x_2,\dots,x_n$ are $2n$ positive real numbers,

\begin{align*} (p_1x_1+\dots +p_nx_n)^2 \leq (p_1+\dots+p_n)(p_1x_1^2+\dots+p_nx_n^2) \end{align*}

b. If $a,b,c$ are positive numbers

\begin{align*} (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq 9a^2b^2c^2 \end{align*}

c. If $x_k,y_k,k=1,2,\dots, n$ are positive numbers,

\begin{align*} \sum_{k=1}^n x_ky_k \leq \left(\sum_{k=1}^n kx_k^2 \right)^{1/2} \left(\sum_{k=1}^n y_k^2/k \right)^{1/2} \end{align*}

d. If $a_k,b_k,c_k,k=1,2,\dots, n$ are positive numbers,

\begin{align*} \left( \sum_{k=1}^n a_k b_k c_k \right)^4 \leq \left(\sum_{k=1}^n a_k^4 \right)\left(\sum_{k=1}^n b_k^4 \right)\left(\sum_{k=1}^n c_k^2 \right)^2 \end{align*}

e. If $C_k = {n \choose k}$ for $n>2$, $1 \leq k \leq n$,

\begin{align*} \sum_{k=1}^n \sqrt{C_k} \leq \sqrt{n(2^n-1)} \end{align*}

### Solution

Using Cauchy Schwarz inequality

a. We have

\begin{align*} \left(\sqrt{p_1}\sqrt{p_1}x_1 + \dots + \sqrt{p_n}\sqrt{p_n}x_n\right)^2 \\ \leq \left(\sqrt{p_1}^2+ \dots + \sqrt{p_n}^2\right)\left((\sqrt{p_1}x_1)^2 + \dots + (\sqrt{p_n}x_n)^2\right) \\ \implies (p_1x_1+\dots +p_nx_n)^2 \leq (p_1+\dots+p_n)(p_1x_1^2+\dots+p_nx_n^2) \end{align*}

b. We have

\begin{align*} \left((a\sqrt{b})^2 + (b\sqrt{c})^2 + (c\sqrt{a})^2\right) \left((c\sqrt{b})^2 + (a\sqrt{c})^2 + (b\sqrt{a})^2\right) \\\geq \left(a\sqrt{b}\cdot c\sqrt{b} + b\sqrt{c}\cdot a\sqrt{c} + c\sqrt{a} \cdot b\sqrt{a} \right)^2 \\ \implies (a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \geq (3abc)^2 = 9a^2b^2c^2 \end{align*}

c. We have

\begin{align*} \left( \sum_{k=1}^n (\sqrt{k}x_k)^2 \right)\left( \sum_{k=1}^n (y_k/\sqrt{k})^2 \right) &\geq \left(\sum_{k=1}^n \sqrt{k}x_k \cdot y_k/\sqrt{k} \right)^2\\ \implies \left(\sum_{k=1}^n kx_k^2 \right)^{1/2} \left(\sum_{k=1}^n y_k^2/k \right)^{1/2} &\geq \sum_{k=1}^n x_ky_k \end{align*}

d. We have

\begin{align*} \left(\sum_{k=1}^n a_k b_k c_k \right)^2 &\leq \left(\sum_{k=1}^n a_k^2 b_k^2 \right) \left(\sum_{k=1}^n c_k^2 \right) \\ \implies \left(\sum_{k=1}^n a_k b_k c_k \right)^2 &\leq \left(\sum_{k=1}^n a_k^4 \right)^{1/2}\left(\sum_{k=1}^n b_k^4 \right)^{1/2} \left(\sum_{k=1}^n c_k^2 \right) \\ \implies \left( \sum_{k=1}^n a_k b_k c_k \right)^4 &\leq \left(\sum_{k=1}^n a_k^4 \right)\left(\sum_{k=1}^n b_k^4 \right)\left(\sum_{k=1}^n c_k^2 \right)^2 \end{align*}

e. We have

\begin{align*} \left( \sum_{k=1}^n 1 \cdot \sqrt{C_k} \right)^2 &\leq \left(\sum_{k=1}^n 1^2 \right) \left(\sum_{k=1}^n C_k \right) \\ \implies \sum_{k=1}^n \sqrt{C_k} &\leq \sqrt{n(2^n-1)} \end{align*}

## Problem 7.3.8

For $n$ a positive integer, let $(a_1, a_2,\dots, a_n)$ and $(b_1, b_2,\dots, b_n)$ be two (not necessarily distinct) permutations of $(1,2,\dots,n)$. Find sharp lower and upper bounds for $a_1b_1+ \dots+a_nb_n$.

### Solution

We have

\begin{align*} a_1b_1+ \dots+a_nb_n \leq \left(\sum_{k=1}^n k^2 \right)^{1/2} \left(\sum_{k=1}^n k^2 \right)^{1/2} = \frac{n(n+1)(2n+1)}{6} \end{align*}

Using the rearrangement inequality we have

\begin{align*} a_1b_1+ \dots+a_nb_n \\ \geq 1\cdot n + \dots + 1\cdot n &= \sum_{k=1}^n k(n+1-k)\\ &= \frac{n(n+1)}{2}\left(n+1 - \frac{2n+1}{3}\right) \\ & = \frac{n(n+1)(n+2)}{6} \end{align*}

## Problem 7.3.9

If $a,b,c,d$ are positive numbers such that $c^2+d^2=(a^2+b^2)^3$ prove that

\begin{align*} \frac{a^3}{c} + \frac{b^3}{d} \geq 1 \end{align*}

with equality if and only if $ad=bc$.

### Solution

We have

\begin{align*} (a^2+b^2)^3 &= c^2+d^2 \\ \implies (a^2+b^2)^4 = (a^2+b^2)(c^2+d^2) &\geq (ac + bd)^2 \\ \implies (a^2+b^2)^2 &\geq (ac + bd) \end{align*}

We also have

\begin{align*} \left( \frac{a^3}{c} + \frac{b^3}{d} \right) (ac+bd) &\geq (a^2+b^2)^2 \geq (ac+bd) \\ \implies \left( \frac{a^3}{c} + \frac{b^3}{d} \right) &\geq 1 \end{align*}

## Problem 7.3.10

Let $P$ be a point in the interior of triangle $ABC$, and let $r_1,r_2,r_3$ denote the distances from P to the sides $a_1,a_2, a_3$ of the triangle respectively. Use Cauchy-Schwarz inequality to show that the minimum value of

\begin{align*} \frac{a_1}{r_1} + \frac{a_2}{r_2} + \frac{a_3}{r_3} \end{align*}

occur when $P$ is at the incenter of triangle ABC.

### Solution

We have

\begin{align*} \left(\left(\sqrt{\frac{a_1}{r_1}}\right)^2+ \left(\sqrt{\frac{a_2}{r_2}}\right)^2 + \left(\sqrt{\frac{a_3}{r_3}}\right)^2 \right) \cdot \\ \left((\sqrt{a_1r_1})^2+ (\sqrt{a_2r_2})^2+ (\sqrt{a_3r_3})^2 \right) \\ \geq (a_1+a_2+a_3)^2 \\ \implies \frac{a_1}{r_1} + \frac{a_2}{r_2} + \frac{a_3}{r_3} \geq \frac{4s^2}{2\Delta} \\ \implies \frac{a_1}{r_1} + \frac{a_2}{r_2} + \frac{a_3}{r_3} \geq \frac{2s}{r} = \frac{a_1}{r} + \frac{a_2}{r} + \frac{a_3}{r} \end{align*}

where $r$ is the inradius, $\Delta$ is the area of $ABC$ and $s$ is the semi-perimeter.