# 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 $x+y>1$?

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

-
+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.

-
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

I will prove it like this:

$\frac{x+1}{2} > (x\times 1)^{\frac{1}{2}}$ --(a) by theorem of mean

and $\frac{y+1}{2} > (y\times 1)^{\frac{1}{2}}$ ---(b)

$\frac{x+y+2}{2} > x^{\frac{1}{2}} + y^{\frac{1}{2}}$
$(x+y) > \frac{(x+y+2)}{2}$
$x+y > 2 > 1$