question
Let x, y and z be strictly positive real numbers. Prove that the inequality holds:
$$\frac{x^5+4x^2+3x+8}{(y+1)^2}+\frac{y^5+4y^2+3y+8}{(z+1)^2}+\frac{z^5+4z^2+3z+8}{(x+1)^2} \ge12$$
my idea
The first thought i had was using Bergstrom inequality, but then i realised i can't turn the fractions. I also tried using CBS but with no result.
The only thing i got and i think might be abit useful is using AM-GM:
$x^5+4x^2+3x\geq 3* x^3*\sqrt{12}$
The radical is of 3 order.
I dont know what else i shoul do. Hope one of you can help me. Thank you!
Best Answer
We prove the inequality by using the rearrangement inequality
We observe that
Then, WOLG, we assume that $x \ge y \ge z$, by using the rearrangement inequality, we have
$$\text{LHS} =\sum_{x,y,z} \frac{x^5+4x^2+3x+8}{(\color{red}{y}+1)^2} \ge \sum_{x,y,z} \frac{x^5+4x^2+3x+8}{(\color{red}{x}+1)^2} \tag{1}$$
And it easy to prove that $$\frac{x^5+4x^2+3x+8}{(x+1)^2} - 4 = \frac{(x-1)^2 (x^3+2x^2+3x+4)}{(x+1)^2} \ge0 \tag{2}$$
From $(1),(2)$, we deduce that $\text{LHS} \ge 12$ and the equality occurs if and only if $x=y=z=1$.