<s>[INST] 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*}
 [/INST] $f(x) = 1/\sqrt{x}$ is strictly convex on $(0, \infty)$, so that
$$
f\left( \frac{x+y}{2}\right) < \frac 12 \left( f(x) + f(y) \right) \, ,
$$
for $0 < x < y$, this gives the stronger inequality
$$
\frac{1}{\sqrt{x+y}}<\frac{1}{2\sqrt 2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right) \, .
$$
</s>