Is the inequality true for all $n\geq 2$?

Question

Let $x,y,z>0$. I am wondering if the following inequality is true?

$$\sum_{cyc}\frac{x^n}{y^2+yz+z^2}\geq\frac{x^{2n-2}+y^{2n-2}+z^{2n-2}}{x^n+y^n+z^n},\qquad n\geq 2$$

If not, is it known for which $n$ it is true?

$\displaystyle(1)\qquad\sum_{cyc}\dfrac{x^2}{y^2+yz+z^2}\overset{AGM}{\geq} \frac{2}{3}\sum_{cyc}\frac{x^2}{y^2+z^2}\overset{Nesbitt}{\geq}1=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}$

$\displaystyle(2)\qquad\sum_{cyc}\dfrac{x^3}{y^2+yz+z^2}=\sum_{cyc}\dfrac{3x^4}{3xy^2+3xyz+3xz^2}\overset{AGM}{\geq}\sum_{cyc}\frac{3x^4}{3x^3+3y^3+3z^3}=\frac{x^4+y^4+z^4}{x^3+y^3+z^3}$

$\displaystyle(3)\qquad\sum_{cyc}\dfrac{x^4}{y^2+yz+z^2}\overset{AGM}{\geq}\frac{2}{3}\sum_{cyc}\dfrac{x^4}{y^2+z^2}\overset{(*)}{\geq}\frac{x^6+y^6+z^6}{x^4+y^4+z^4}$

$(*)\iff\forall$ $a,b,c>0$, $\,\,\displaystyle \frac{2}{3}\sum_{cyc}\dfrac{a^2}{b+c}\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}$

$\iff \displaystyle \frac{2}{3}\left(\sum_{cyc}\dfrac{a^2}{b+c}-\frac{a+b+c}{2}\right)\geq\frac{a^3+b^3+c^3}{a^2+b^2+c^2}-\frac{a+b+c}{3}$

$\iff\displaystyle\frac{(a+b+c)}{3(a+b)(b+c)(c+a)}\sum_{cyc}(a+b)(a-b)^2\geq\frac{1}{3(a^2+b^2+c^2)}\sum_{cyc}(a+b)(a-b)^2$

$\iff\displaystyle\left((a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)\right)\sum_{cyc}(a+b)(a-b)^2\geq0$

where

$$(a+b+c)(a^2+b^2+c^2)-(a+b)(b+c)(c+a)$$ $$\ge(a+b+c)(ab+bc+ca)-(a+b)(b+c)(c+a)=abc>0$$

Done!

Answer 1

It's wrong for any $n\geq5$.

For $n=5$ try $x=1.1$ and $y=z=1$.

For $n=4$ and $n=3$ it's true.

Answer 2

A proof for $n=4$.

We need to prove that: $$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\frac{x^6+y^6+z^6}{x^4+y^4+z^4}.$$ Now, by AM-GM $$\sum_{cyc}\frac{x^4}{y^2+yz+z^2}\geq\sum_{cyc}\frac{x^4}{y^2+\frac{y^2+z^2}{2}+z^2}=\frac{2}{3}\sum_{cyc}\frac{x^4}{y^2+z^2}.$$ Id est, it's enough to prove that $$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{3}{2}\cdot\frac{x^3+y^3+z^3}{x^2+y^2+z^2}$$ for positives $x$, $y$ and $z$.

Now, by C-S $$\sum_{cyc}\frac{x^2}{y+z}=\sum_{cyc}\frac{x^6}{x^4y+x^4z}\geq\frac{(x^3+y^3+z^3)^2}{\sum\limits_{cyc}(x^4y+x^4z)}.$$ Thus, it's enough to prove that $$2(x^3+y^3+z^3)(x^2+y^2+z^2)\geq3\sum\limits_{cyc}(x^4y+x^4z)$$ or $$\sum_{cyc}(2x^5-3x^4y-3x^4z+2x^3y^2+2x^3z^2)\geq0$$ or
$$\sum_{cyc}(x^5-3x^4y+2x^3y^2+2x^2y^3-3xy^4+y^5)\geq0$$ or $$\sum_{cyc}(x+y)(x-y)^4\geq0$$ and we are done!