Mathematical Reflections 2016, Issue 5 - Problem S385

Problem:
Let $a,b,c$ be positive real numbers. Prove that $$\dfrac{1}{a^3+8abc}+\dfrac{1}{b^3+8abc}+\dfrac{1}{c^3+8abc} \leq \dfrac{1}{3abc}.$$

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

Solution:
The given inequality can be written as $$\dfrac{1}{9abc}-\dfrac{1}{a^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{b^3+8abc}+\dfrac{1}{9abc}-\dfrac{1}{c^3+8abc} \geq 0,$$ i.e. $$\dfrac{a^2-bc}{a^2+8bc}+\dfrac{b^2-ca}{b^2+8ca}+\dfrac{c^2-ab}{c^2+8ab} \geq 0.$$
Let $x=\dfrac{bc}{a^2}$, $y=\dfrac{ca}{b^2}$, $z=\dfrac{ab}{c^2}$. Then, we have $xyz=1$ and we want to prove that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0.$$ Observe that $$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z}=\dfrac{3(-64xyz+16(xy+yz+zx)+5(x+y+z)+1)}{(1+8x)(1+8y)(1+8z)}.$$ Since $xyz=1$, then by AM-GM Inequality we have $x+y+z \geq 3$ and $xy+yz+zx \geq 3$, so $$-64xyz+16(xy+yz+zx)+5(x+y+z)+1 \geq -63+16\cdot3+5\cdot3=0$$ and we get
$$\dfrac{1-x}{1+8x}+\dfrac{1-y}{1+8y}+\dfrac{1-z}{1+8z} \geq 0,$$ as we wanted to prove.