Let x,y,z such a positive integers we can also prove it for real numbers with $x>y$ Prove that: $$\frac{(x+y+z)^3\sqrt{xy+y^2}}{(\sqrt{x+y}-\sqrt{x-y})(\sqrt{x}+\sqrt{y})}\geq \frac{27(\sqrt{x}-\sqrt{y})xyz}{\sqrt{x-y}}$$ I tried to use AM GM
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3Presumably $x>y$ or otherwise you are going to have a problem with $\sqrt{x-y}$? – Thomas Andrews Feb 19 '16 at 13:05
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@user315918: you could maybe also accept these suitable answers : (1), (2), (3), (4), (5), (6), (7), (8). Many thanks :-) ! – Watson Jun 20 '16 at 09:14
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And mabye these ones : (9), (10), (11), (12), (13), (14). This would be great! – Watson Jun 20 '16 at 09:16
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we have $$(x+y+z)^3\geq 27xyz$$ by $$AM-GM$$ by multiplying the given inequality by $$\frac{\sqrt{y}\sqrt{x+y}}{\sqrt{x+y}-\sqrt{x-y}}>0$$ we have to prove that $$\frac{27xyz\sqrt{y}\sqrt{x+y}}{\sqrt{x-y}-\sqrt{x-y}}\geq 27xyz\sqrt{x-y}$$ but this is true since $$\sqrt{y}\sqrt{x+y}+x-y\geq \sqrt{x+y}\sqrt{x-y}$$ this is true since after squaring we get $$y(x+y)+(x-y)(2\sqrt{y}\sqrt{x+y}-2y)\geq 0$$ since $$2\sqrt {y}\sqrt{x+y}\geq 2y$$ this is true since $$x>0$$
Dr. Sonnhard Graubner
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I may be nitpicking, but there are some words you typed wrong. Such as poove $\rightarrow$ prove, and t5rue $\rightarrow true. Otherwise, your answer is nice. – S.C.B. Feb 19 '16 at 14:45