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Let $z \geq y \geq 1$. Show that

$$ \sqrt{\frac{y}{1+z}} + \sqrt{\frac{z}{1+y}} + \sqrt{\frac{1}{y+z}} > 2 $$

These are actually the last steps of a bigger inequality, but I can't think of a nice way to prove it (no differentiation please).

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Equivalently, we may prove the inequality $$ \sum_{cyc}\sqrt{\frac{x}{1-x}}\geq 2 \tag{1}$$ under the constraints $x,y,z\geq 0$ and $x+y+z=1$. That follows from: $$ \forall x\in(0,1),\quad \sqrt{\frac{x}{1-x}}\geq 2x \tag{2}$$ that is equivalent to: $$ \forall x\in(0,1),\quad 4x(1-x)\leq 1 \tag{3} $$ that is a straightforward consequence of the AM-GM inequality.

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  • $\begingroup$ Thank you. I had obviously not chosen the right constraint by setting $x = 1$. $\endgroup$ – nabla Jun 14 '15 at 21:55

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