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.

Here's a solution using Jensen's inequality.
To get rid of the constraint, let's set
$$
x = 2b/a \quad y = 2c/b \quad z = 2a/c
$$
the inequality becomes
$$
\sqrt{\frac a {a + 2b}} + \sqrt{\frac b {b + 2c}} + \sqrt{\frac c {c + 2a}} < 2
$$
Since $\sqrt t$ is a concave function, applying Jensen's inequality we get
$$
\begin{align}
\sum_{cyc}\sqrt{\frac a {a + 2b}} &= 
\sum_{cyc} \frac {a + c} {2(a + b + c)} \sqrt{\frac{4(a + b + c)^2}{(a + c)^2} \frac a {a + 2b}}\\
&\leq \sqrt{\sum_{cyc} \frac{2a(a + b + c)}{(a + c)(a + 2b)}}
\end{align}
$$
Now, it suffices to expand the expression $E$ inside the last square root. 
If we call $N$ its numerator and $D$ its denominator then $4D - N$ results in a sum of monomials in $a, b$ and $c$ with positive coefficients. Therefore
$$
E = \frac N D < 4
$$
and the inequality is proved.
