Question
Let $x$, $y$, and $z$ be positive real numbers. Prove the inequality:
A. $\frac{x^4+x^2+7}{y+2}+\frac{y^4+y^2+7}{z+2}+\frac{z^4+z^2+ 7}{x+2}>=x+y+z+6$
B.
$\frac{x^4+x^2+7}{(y+2)^2}+\frac{y^4+y^2+7}{(z+2)^2}+\frac{z^4+z^2+ 7}{(x+2)^2}>=3$
idea
So we have to show that
$\frac{x^4+x^2+7}{y+2}>=y+2$
And do the same thing to others.
We will be able to demonstrate the jnequality if we show this.
I thought of the inequality of means or the Minkovsky inequality but none of them works for me.
Also, we can suppose without lose of generality that $x<=y<=z$.
I dont know what to do forward. Hope one of you can help me. Thank you!
Best Answer
By AM-GM $$x^4+x^2+7-(x+2)^2=x^4-4x+3\geq4\sqrt[4]{x^4\cdot1^3}-4x=0.$$
Thus, for A it remains to prove that $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq a+b+c,$$ where $x+2=a$, $y+2=b$ and $z+2=c$,
which is true by C-S: $$\sum_{cyc}\frac{a^2}{b}\geq\frac{(a+b+c)^2}{a+b+c}=a+b+c$$ or by AM-GM: $$\sum_{a^2}{b}=\sum_{cyc}\left(\frac{a^2}{b}+b-b\right)\geq\sum_{cyc}\left(2\sqrt{\frac{a^2}{b}\cdot b}-b\right)=\sum_{cyc}(2a-b)=a+b+c.$$ Also, it's true by Rearrangement and there are many other ways to a proof.
For $B$ it remains to prove that $$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq3,$$ which is true by AM-GM.