TITLE: How prove this $\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\le\frac{2}{\sqrt{1+\sqrt{xy}}}$
QUESTION [9 upvotes]: Let $x,y>0$ and $xy\le 1$. Show that
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}\le\dfrac{2}{\sqrt{1+\sqrt{xy}}}.$$
This inequality have same  follow methods?
I saw this.
Let $x,y>0, xy\le 1$. 
$$\dfrac{1}{1+x}+\dfrac{1}{1+y}\le\dfrac{2}{1+\sqrt{xy}},$$
because we have
\begin{align}
\dfrac{1}{1+x}+\dfrac{1}{1+y}-\dfrac{2}{1+\sqrt{xy}}&=\left(\dfrac{1}{1+x}-\dfrac{1}{1+\sqrt{xy}}\right)+\left(\dfrac{1}{1+y}-\dfrac{1}{1+\sqrt{xy}}\right)\\
&=\dfrac{(\sqrt{x}-\sqrt{y})^2(\sqrt{xy}-1)}{(1+x)(1+y)(1+\sqrt{xy})}\le 0
\end{align}
Thank you everyone, or have other nice methods?

REPLY [5 votes]: As you show
$$\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}},$$
it follows that
$$(\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}})^2\le 2 (\frac{1}{1+x}+\frac{1}{1+y})\le \frac{4}{1+\sqrt{xy}}.$$