Imad Zak's Cyclic Inequality in Three Variables

Solution 1

By the AM-GM inequality,

$\displaystyle \frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{ab+bc+ca}{a^2+b^2+c^2}\ge 2.$

Then

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

which is exactly

$\displaystyle \frac{(a+b+c)^2}{ab+bc+ca} + \frac{ab+bc+ca}{a^2+b^2+c^2}\ge 4.$

Equality for $a=b=c.$

Solution 2

Set $p=a+b+c,\,$ $q=ab+bc+ca\,$ and note that $\displaystyle \sum_{cycl}a^2=p^2-2q,\,$ which shows that $p^2-2q\ge 0\,$ and reduces the required inequality to $\displaystyle \frac{p^2}{q}+\frac{q}{p^2-2q}\ge 4,\,$ or

$p^2(p^2-2q)+q^2\ge 4q(p^2-2q),$

same as $p^4-6p^2q+9q^2=(p^2-3q)^2\ge 0.\,$ Equality when $p^2-3q=0,\,$ i.e., $a=b=c.$

Acknowledgment

This problem has been posted at the mathematical inequalities facebook group by Imad Zak. Solution 1 is by Ravi Prakash; Solution 2 is by Nguyen Lam Hong.