Show inequality for positive real numbers If $x,y$ are positive real numbers then we have that $$ \frac{1}{\sqrt{x+y}}<\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}$$ right? But how can we show that?
I have tried the following but I don't think that this is the way we should go.
\begin{align*}\frac{1}{\sqrt{x + y}} < \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}}=\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}} & \iff \left(\frac{1}{\sqrt{x + y}}\right)^2 < \left(\frac{\sqrt{y}+\sqrt{x}}{\sqrt{xy}}\right)^2 \\ & \iff \frac{1}{x + y} <\frac{x+y+2\sqrt{xy}}{xy} \\ & \iff  xy<(x+y)(x+y+2\sqrt{xy}) \\ & \iff xy<x^2+xy+2x\sqrt{xy}+xy+y^2+2y\sqrt{xy}\\ & \iff 0<x^2+2(x+y)\sqrt{xy}+xy+y^2\end{align*}

The stronger inequality in the answer by Martin R can be proved by applying $a^2+b^2>2ab,$ for $0<a<b,$ twice. For $0<u<v$ we get $${1\over \sqrt{u^4+v^4}}<{1\over \sqrt{2}uv}={1\over 2\sqrt{2}}2{1\over u}{1\over v}<{1\over 2\sqrt{2}}\left ({1\over u^2}+{1\over v^2}\right )$$ Now plug in $x=u^4$ and $y=v^4.$
