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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!

2 Answers 2

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

$2\sqrt{xy}

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

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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 $x+y\leq \sqrt x+\sqrt y$