# Inequality's root

Given $\sqrt x + \sqrt y < x+y$, prove that $x+y>1$.

Havnt been able to try this yet, found it online,

any help is appreciated thanks!

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If $\sqrt{x}+\sqrt{y}<x+y$, then after squaring both sides we have $x+2\sqrt{xy}+y<x^2+2xy+y^2$. Rearrange to isolate the square root:

$$2\sqrt{xy}<x^2+2xy+y^2-x-y=(x+y)^2-(x+y)=(x+y)(x+y-1)\;.$$

Can you see why this implies that $x+y>1$?

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yes thanks alot – fosho Jul 11 '12 at 19:56

If $x+y\leq 1$ then $0\leq x,y \leq 1$ (since $x,y \geq 0$). It follows that $x\leq \sqrt x$ and $y\leq \sqrt {y}$ which implies that $x+y\leq \sqrt x+\sqrt y$

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+1 for such a nice solution! – Belgi Jul 11 '12 at 20:25
@Belgi Thank you. – azarel Jul 11 '12 at 20:45
are you sure this is correct? – fosho Jul 19 '12 at 17:44

$1 < \frac{\sqrt x +2\sqrt xy +\sqrt y}{\sqrt x +\sqrt y} =\sqrt x+\sqrt y$

Now if you square on both side you get the result.

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Not true. $(\sqrt{x}+\sqrt{y})^2 = x + 2\sqrt{xy} + y$. – marty cohen Jul 11 '12 at 22:58
@martycohen : I meant to square the two sides of $1< \sqrt x + \sqrt y$ – Theorem Jul 12 '12 at 3:34