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Let $x,y,z>0$ such that $xyz=1$. Show that: $$\dfrac{x^3+1}{\sqrt{x^4+y+z}}+\dfrac{y^3+1}{\sqrt{y^4+z+x}}+\dfrac{z^3+1}{\sqrt{z^4+x+y}}\geq2\sqrt{xy+yz+zx}$$

I've tried many things but all failed. Please help. Thank you.

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  • $\begingroup$ This looks like fun $\endgroup$ Jan 23, 2014 at 8:52

1 Answer 1

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proof: since \begin{align*}2\sqrt{(x^4+y+z)(xy+zx+yz)}&=2\sqrt{[x^4+xyz(y+z)][xy+yz+xz]}\\ &=2\sqrt{(x^3+y^2z+yz^2)(x^2y+ x^2z+xyz)} \end{align*} Use AM-GM inequality we have $$2\sqrt{(x^3+y^2z+yz^2)(x^2y+x^2z+xyz)}\le x^3+y^2z+yz^2+x^2y+x^2z+xyz=(x+y+z)(x^2+yz)=\dfrac{(x+y+z)(x^3+1)}{x}$$ so $$\dfrac{x^3+1}{\sqrt{x^4+y+z}}\ge\dfrac{2x\sqrt{xy+yz+xz}}{x+y+z}$$ so $$\sum_{cyc}\dfrac{x^3+1}{\sqrt{x^4+y+z}}\ge\sum_{cyc}\dfrac{2x\sqrt{xy+yz+xz}}{x+y+z}=2\sqrt{xy+yz+xz}$$

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