Suppose $xyz=8$, try to prove that $\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2$ Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always less than 1 for $t>0$.
I'm just wondering if we can have some elegant simple prove. I guess we should use Jensen's inequality. Thanks.
Suppose $x,y,z\in R^+$ and $xyz=8$, try to prove that $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2.$$
Please note that the usual AM-GM inequality may not do its trick here as the equality is rather hold on the boundary.
Thanks.

Let $f(x,y,z)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ and WOLOG assume that $x\geq y\geq z$ so we have $x\geq 2$. Consider the following two cases:


*

*$yz\geq 1$ : In this case we have $z\geq1/y$ and thus
$$f(x,y,z)\leq f(x,y,1/y)\\ \leq\sqrt{3\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{y}{1+y}\right)}\\
= \sqrt{3\left(\frac{1}{1+x}+1\right)}\\
\leq \sqrt{3\left(\frac{1}{3}+1\right)}=2.$$ Equation in this case cannot be attained as we cannot have $z=1/y$ and $x=2$ simultaneously.

*$yz\leq 1$: Fix the product $yz=p^2$. Since we assumed $y\geq z$ we must have $y\geq p$. Furthermore, we have $p\leq 1$ for which the function $\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ would be decreasing with respect to $y$.  Therefore, the mentioned function is minimized at $y=z=p$, i.e., $$\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}\leq\frac{2}{\sqrt{1+\sqrt{yz}}}\\=\frac{2}{\sqrt{1+\sqrt{8/x}}}.$$ Thus $f(x,y,z)\leq \frac{1}{\sqrt{1+x}}+\frac{2}{\sqrt{1+\sqrt{8/x}}}.$ The RHS is increasing in $x$ so the upper bound as $x\to \infty$ would be $f(x,y,z)<2$.


There might be neater way to show the second case.
