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! 

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 $x+y>1$?
