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 $x+y+z=3$?

Question

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!

Answer 1

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:

1. $y=z=1$, which gives $$(x-1)^2(x^6+6x^5+291x^4-48x^3-270x^2-120x+248)\geq0,$$ which is true by AM-GM;
2. $z\rightarrow0^+$ and $y=1$, which gives: $$x^6+2x^5+138x^4-158x^3+138x+2x+1\geq0,$$ which is true by AM-GM again.

Answer 2

Another way.

We need to prove that: $$\sum_{cyc}\left(\frac{1}{x^2-xy+y^2}-\frac{9}{(x+y+z)^2}\right)+6\sum_{cyc}(1-\sqrt{xy})\geq0$$ or $$\sum_{cyc}\frac{z^2-8x^2-8y^2+11xy+2xz+2yz}{x^2-xy+y^2}+27\sum_{cyc}(x+y-2\sqrt{xy})\geq0$$ or $$\sum_{cyc}\frac{(z-x)(16x-11y+z)-(y-z)(16y-11z+x)}{x^2-xy+y^2}+54\sum_{cyc}(\sqrt{x}-\sqrt{y})^2\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{16y-11z+x}{y^2-yz+z^2}-\frac{16x-11z+y}{x^2-xz+z^2}\right)+\frac{54(x-y)^2}{(\sqrt x+\sqrt y)^2}\geq0$$ or $$\sum_{cyc}(x-y)^2\left(\frac{x^2+y^2+17xy-4z^2-12xz-12yz}{(x^2-xz+z^2)(y^2-yz+z^2)}+\frac{162}{(x+y+z)(\sqrt x+\sqrt y)^2}\right)\geq0$$ and since by C-S $(\sqrt x+\sqrt y)^2\leq2(x+y),$ it's enough to prove that: $$(x^2+y^2+17xy-4z^2-12xz-12yz)(x+y)(x+y+z)+81(x^2-xz+z^2)(y^2-yz+z^2)\geq0$$ or $$81z^4-85(x+y)z^3+(65x^2+49xy+65y^2)z^2-11(x^3+9x^2y+9xy^2+y^3)z+x^4+19x^3y+117x^2y^2+19xy^3+y^4\geq0$$ or $$81z^4-85(x+y)z^3+23(x+y)^2z^2+$$ $$+(42x^2+3xy+42y^2)z^2-11(x^3+9x^2y+9xy^2+y^3)z+x^4+19x^3y+117x^2y^2+19xy^3+y^4\geq0,$$ which is true by AM-GM: $$81z^4-85(x+y)z^3+23(x+y)^2z^2\geq2z^2\sqrt{81z^4\cdot23(x+y)^2z^2}-85(x+y)z^3=$$ $$=\left(18\sqrt{23}-85\right)z^3(x+y)>0$$ and $$(42x^2+3xy+42y^2)z^2-11(x^3+9x^2y+9xy^2+y^2)z+x^4+19x^3y+117x^2y^2+19xy^3+y^4\geq$$ $$\left(2\sqrt{(42x^2+3xy+42y^2)(x^4+19x^3y+117x^2y^2+19xy^3+y^4)}-11(x^3+9x^2y+9xy^2+y^3)\right)z>0.$$