# Imad Zak's Cyclic Inequality in Three Variables II

### Solution 1

Define $\displaystyle f(t)=t+\frac{1}{t+2}.\,$ The function is increasing for $t\ge 0.\,$ Note that $\displaystyle \frac{a^2+b^2+c^2}{ab+bc+ca}\ge 1,\,$ implying

$\displaystyle \frac{ab+bc+ca}{(a+b+c)^2} + \frac{a^2+b^2+c^2}{ab+bc+ca}=f\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)\ge f(1)=\frac{4}{3}.$

### Solution 2

Let $\displaystyle x=\sum_{cycl}a^2,\,$ $\displaystyle p=\sum_{cycl}a\,$ and $\displaystyle q=\sum_{cycl}ab.\,$ Let

$\displaystyle A=\frac{x}{q}+\frac{q}{p^2}=\frac{2}{3}\cdot\frac{x}{q}+\frac{1}{3}\cdot\frac{x}{q}+\frac{q}{p^2}=\frac{2}{3}\cdot\frac{x}{q}+B.$

We know that $\displaystyle \frac{x}{q}\ge 1\,$ and $\displaystyle x\ge\frac{1}{3}p^2,\,$ so that

$B\ge\displaystyle \frac{1}{9}\cdot\frac{p^2}{q}+\frac{q}{p^2}\ge 2\sqrt{\frac{1}{9}\cdot\frac{p^2}{q}\cdot\frac{q}{p^2}}=2\cdot\frac{1}{3},$

by the AM-GM inequality. Thus $\displaystyle A\ge\frac{2}{3}+\frac{2}{3}=\frac{4}{3}.$ Equality at $a=b=c.$

### Solution 3

Let $s=a^2+b^2+c^2,\,$ $t=ab+bc+ca.\,$ Then

$\displaystyle L=\frac{ab+bc+ca}{(a+b+c)^2} + \frac{a^2+b^2+c^2}{ab+bc+ca}=\frac{s}{t}+\frac{t}{s+2t}.$

With $\lambda=\displaystyle \frac{s}{t},\,$ we have

$\displaystyle L=\lambda+\frac{1}{\lambda+2}=\frac{(\lambda+1)^2}{\lambda+2}.$

Since $f'(\lambda)\gt 0\,$ for $\lambda\gt 0,\,$ telling us that $f(\lambda)\,$ is an increasing function, so we must have $L(\lambda)\ge L(\lambda^{*}),\,$ where $\lambda^{*}\,$ is the minimum of the expression $\displaystyle \frac{s}{t}=\frac{a^2+b^2+c^2}{ab+bc+ca},\,$ which is $1.\,$ Thus we have

$\displaystyle L(\lambda)\ge\frac{(1+1)^2}{1+2}=\frac{4}{3}.$

### Solution 5

$\displaystyle \frac{ab+bc+ca}{(a+b+c)^2} + \frac{a^2+b^2+c^2}{ab+bc+ca}\ge\frac{p^2}{3q}+\frac{q}{p^2}=\lambda+\frac{1}{3\lambda},$

where $\lambda\ge 1.\,$ Now, $(\lambda-1)(3\lambda-1)\ge 0,\,$ so that $3\lambda^2-4\lambda=1\ge 0\,$ which is equivalent to $\displaystyle \lambda+\frac{1}{3\lambda}\ge\frac{4}{3}.$

### Solution 6

$\displaystyle A=\frac{(a+b+c)^2}{ab+bc+ca}=\frac{a^2+b^2+c^2}{ab+bc+ca}+2.$

So we need to prove that $\displaystyle A-2+\frac{1}{A}\ge\frac{4}{3}.\,$ Solving the quadratic equation $3A^2-10A+3=0\,$ give two roots: $3\,$ and $\displaystyle \frac{1}{3}.\,$ Thus, it's enough to show that $A\ge 3.\,$ This follows from the well known and easily proved inequality $3(ab+bc+ca)\le (a+b+c)^2.$

### Solution 7

We already know that, $\displaystyle \frac{(a+b+c)^2}{ab+bc+ca} + \frac{ab+bc+ca}{a^2+b^2+c^2}\ge 4.\,$ If $\displaystyle X=\frac{(a+b+c)^2}{ab+bc+ca}\,$ and $\displaystyle \frac{ab+bc+ca}{a^2+b^2+c^2}=Y\,$ then $X+Y\ge 4\,$ and we have to estimate

$\displaystyle \frac{1}{X}+\frac{1}{Y}=\frac{X+Y}{XY}\ge\frac{4}{XY}\ge\frac{4}{3},$

because

$\displaystyle XY= \frac{(a+b+c)^2}{a^2+b^2+c^2}\le 3.$

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Imad Zak. Solution 1 is by Daniel Dan; Solution 2 is by Imad Zak; Solution 3 is by David Holden, Solution 4 is by Nassim Nicholas Taleb; Solution 5 is by Aziz Atta; Solution 6 has been submitted anonymously on twitter. The original post is at the mathematical inequalities facebook group.

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