How prove this $\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}\le\frac{2}{\sqrt{1+\sqrt{xy}}}$

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?

• there is considerable difference between the two versions of the question.Did you really mean it? – lab bhattacharjee Jun 3 '13 at 8:51
• Please do not use \dfrac in question titles. – J. M. isn't a mathematician Jun 3 '13 at 8:54

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