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*}
Best Answer
In fact, for $x,y>0$, one has $$ x+y>x \Longrightarrow \sqrt{x+y}>\sqrt x \Longrightarrow\frac{1}{\sqrt{x+y}}<\frac1{\sqrt x}<\frac1{\sqrt x}+\frac1{\sqrt y.}$$