Mathematical Reflections 2017, Issue 3 - Problem J412

Problem:
Let $a \geq b \geq c$ be positive real numbers. Prove that
$$(a-b+c)\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}+\dfrac{1}{c+a}\right) \leq \dfrac{1}{2}.$$

Proposed by An Zhenping, Xianyang Normal University, China

Solution:
If $a \geq b \geq c$ are positive real numbers, then $a=c+x+y$ and $b=c+y$, where $x,y \geq 0$. The given inequality is equivalent to
$$(x-c)\left(\dfrac{1}{2c+x+2y}-\dfrac{1}{2c+y}+\dfrac{1}{2c+x+y}\right) \leq \dfrac{1}{2}.$$
This inequality is equivalent to
$$\dfrac{16 c^3 + 24 c^2 y + 12 c y^2 + 2 x^3 + 3 x^2 y + x y^2 + 2 y^3}{(2c+x+2y)(2c+y)(2c+x+y)} \geq 0,$$
which is clearly true.