How do I prove that $\sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge6(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$ given $x,y,z > 0$ and $x+y+z=3$? I tried to apply AM-GM inequality:
$$ \dfrac{1}{x^2-xy+y^2} + (x^2-xy+y^2) \ge 2 \implies \dfrac{1}{x^2-xy+y^2} \ge 2 - (x^2-xy+y^2) $$
Then,
$$ \sum_{cyc}\left(\dfrac{1}{x^2-xy+y^2}\right)+15\ge 21-2(x^2+y^2+z^2)+xy+yz+zx $$
I am left to prove
$$ 21-2(x^2+y^2+z^2)+xy+yz+zx \ge 6\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right) $$
which is equivalent to
$$ 6\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)+2(x^2+y^2+z^2)-(xy+xz+zx) \le 21. $$
Before I attempted the problem, I noticed that the equality case for the inequality to prove is at $x=y=z=1$. While these values would satisfy the inequality above, I found out from WolframAlpha that the max of LHS is greater than $21$ (approximately $22.4545$).
Thanks in advance for the help!

We need to prove that:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}+15\geq6(xy+xz+yz),$$ where $x$, $y$ and $z$ are positives such that $x^2+y^2+z^2=3$.
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $3u^2-2v^2=1$ and by C-S we obtain:
$$\sum_{cyc}\frac{1}{x^4-x^2y^2+y^4}=\sum_{cyc}\frac{z^2}{x^4z^2-x^2y^2z^2+y^4z^2}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^4y^2+x^4z^2-x^2y^2z^2)}=$$
$$=\frac{9u^2}{(x^2+y^2+z^2)(x^2y^2+x^2z^2+y^2z^2)-6x^2y^2z^2}=\frac{3u^2}{3(3u^2-2v^2)(3v^4-2uw^3)-2w^6}$$ and it's enough to prove that $f(w^3)\geq0,$ where
$$f(w^3)=\frac{u^2}{3(3u^2-2v^2)(3v^4-2uw^3)-2w^6}+\frac{5}{(3u^2-2v^2)^2}-\frac{6v^2}{(3u^2-2v^2)^3}.$$
But, $$f'(w^3)=-\frac{u^2(-3(3u^2-2v^2)2u-4w^3)}{(3(3u^2-2v^2)(3v^4-2uw^3)-2w^6)^2}>0,$$ which says that it's enough to prove $f(w^3)\geq0$ or $$\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^4y^2+x^4z^2-x^2y^2z^2)}+\frac{135}{(x^2+y^2+z^2)^2}\geq\frac{162(xy+xz+yz)}{(x^2+y^2+z^2)^3}$$ for the minimal value of $w^3$, which by reasoning like here says us, that it's enough to check two cases:

*

*$y=z=1$, which gives $$(x-1)^2(x^6+6x^5+291x^4-48x^3-270x^2-120x+248)\geq0,$$ which is true by AM-GM;

*$z\rightarrow0^+$ and $y=1$, which gives:
$$x^6+2x^5+138x^4-158x^3+138x+2x+1\geq0,$$ which is true by AM-GM again.

