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I was solving a problem on triangular inequalities and I have ended up with the following inequality required to be proven

$$\sum_{cyc}\frac{x}{2x+y+z}\geq\frac{9\sqrt{3(x+y+z)xyz}}{4(x+y+z)^2},$$

How do I prove this? I only know CS inequality and AM GM HM inequality. CAn it be proved using these?

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closed as off-topic by Saad, Christopher, Wouter, Theoretical Economist, Adrian Keister Oct 5 '18 at 16:18

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By C-S $$\sum_{cyc}\frac{x}{2x+y+z}=\sum_{cyc}\frac{x^2}{2x^2+xy+xz}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x^2+2xy)}.$$ Thus, it's enough to prove that $$\frac{(x+y+z)^2}{\sum\limits_{cyc}(2x^2+2xy)}\geq\frac{9\sqrt{3(x+y+z)xyz}}{4(x+y+z)^2}$$ or $$\frac{(x+y+z)^4}{\left(\sum\limits_{cyc}(x^2+xy)\right)^2}\geq\frac{243xyz}{4(x+y+z)^3}$$ or $$4(x+y+z)^7\geq243xyz\left(\sum\limits_{cyc}(x^2+xy)\right)^2.$$ Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

By AM-GM easy to show that $u^2\geq v^2$ and $v^4\geq uw^3$.

Thus, it's enough to prove that $$4u^8\geq(3u^2-v^2)^2v^4$$ or $$2u^4\geq(3u^2-v^2)v^2$$ or $$(u^2-v^2)(2u^2-v^2)\geq0$$ and we are done!

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    $\begingroup$ How did the second part of your first statement come up? $\endgroup$ – saisanjeev Oct 5 '18 at 11:03
  • $\begingroup$ @saisanjeev I added something. See now. $\endgroup$ – Michael Rozenberg Oct 5 '18 at 12:54

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