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 $x+y\leq 1$ 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 $x+y\leq \sqrt x+\sqrt y$
