Mathematical Reflections 2017, Issue 1 - Problem S397

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}+\dfrac{3(ab+bc+ca)}{2(a+b+c)} \geq a+b+c.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
The given inequality is equivalent to $$(a+b+c)\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)+\dfrac{3}{2}(ab+bc+ca) \geq (a+b+c)^2,$$ i.e.

$$\dfrac{a^2c}{a+b}+\dfrac{b^2a}{b+c}+\dfrac{c^2b}{c+a} \geq \dfrac{1}{2}(ab+bc+ca) \qquad (1)$$

By the AM-GM Inequality, we have $$\dfrac{2a^2c}{a+b}+\dfrac{c(a+b)}{2} \geq 2ca,$$ $$\dfrac{2b^2a}{b+c}+\dfrac{a(b+c)}{2} \geq 2ab,$$ $$\dfrac{2c^2b}{c+a}+\dfrac{b(c+a)}{2} \geq 2bc.$$ Adding these three inequalities, we get inequality (1). The equality holds if and only if $a=b=c$.