Prove an inequality, knowing that $x,y,z$ are strictly positive real numbers

Question

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!

Answer 1

By AM-GM twice: $$\sum_{cyc}\tfrac{x^5+4x^2+3x+8}{(y+1)^2}=\sum_{cyc}\tfrac{x^5+4-5x+4(x+1)^2}{(y+1)^2}\geq\sum_{cyc}\tfrac{4(x+1)^2}{(y+1)^2}\geq12.$$

We used the following AM-GM: $$x^5+4-5x=x^5+4\cdot1-5x\geq5\sqrt[5]{x^5\cdot1^4}-5x=0$$ and $$\sum_{cyc}\frac{4(x+1)^2}{(y+1)^2}\geq4\cdot3\sqrt[3]{\prod_{cyc}\frac{(x+1)^2}{(y+1)^2}}=12.$$

Answer 2

We prove the inequality by using the rearrangement inequality

We observe that

• the numerator is an increasing function. Indeed, let's denote $$f(x) = x^5+4x^2+3x+8$$ then $$f'(x) = 5x^4+8x+3 >0 \hspace{1cm} \text{for }x>0$$
• the denominator $x \mapsto(x+1)^2$ is an increasing function, then the function $x \mapsto \frac{1}{(x+1)^2}$ is decreasing.

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 $x=y=z=1$.