# A Cyclic Inequality in Three Variables V

### Solution

We have $(c-a)(c-b)(b-a)\ge 0\,$ so that

$ab^2+bc^2+ca^2\ge a^2b+b^2c+c^2a.$

Hence, suffice it to prove that

$\displaystyle\frac{ab^2+bc^2+ca^2+a^2b+b^2c+c^2a}{abc}\ge\frac{2\sqrt{3(a^2+b^2+c^2)}(a+b+c)}{ab+bc+ca}.$

This is equivalent to

$\displaystyle (a+b+c)(ab+bc+ca)\ge 3abc+\frac{2abc\sqrt{3(a^2+b^2+c^2)}(a+b+c)}{ab+bc+ca}.$

Assume, WLOG, $a+b+c=3.\,$ Then $ab+bc+ca=3(1-t^2),\,$ $0\le t\lt 1\,$ and $\max(abc)=(1-t)^2(1+2t).\,$ Suffice it to show that

$\displaystyle t^3+2t^2+2t+1\ge (1+2t)\sqrt{1+2t^2}$

which is equivalent to

$\displaystyle \frac{t^4(t+2)^2}{(1+2t)^2}+\frac{2t^2(t+2)}{1+2t}\ge 2t^2,$

i.e.,

$\displaystyle \frac{t^2(t+2)^2}{(1+2t)^2}+\frac{2(t+2)}{1+2t}\ge 2.$

But, for $t\in [0,1),\,$ $\displaystyle\frac{2(t+2)}{1+2t}\ge 2\,$ and, clearly, $\displaystyle\frac{t^2(t+2)^2}{(1+2t)^2}\ge 0.$

### Acknowledgment

Leo Giugiuc has kindly communicated to me the problem above, along with the solution of his. The problem has been previously posted by Cooper Carpenter at the mathematical inequalities facebook group. The problem is credited to Le Viet Hung.