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.

It can be proved by contradiction. Set $a=\sqrt{1/(1+x)}$, $b=\sqrt{1/(1+y)}$ and $c=\sqrt{1/(1+z)}$. We have $$8a^2b^2c^2=(1-a^2)(1-b^2)(1-c^2).$$ We need to prove $a+b+c<2$. Assume that $a+b+c\geq 2$, we have
$$1-a^2=(1+a)(1-a)< 2(b+c-1)$$
Since $(1-b)(1-c)>0$, we have $$1-a^2<2bc$$
Then, we have
$$(1-a^2)(1-b^2)(1-c^2)<8a^2b^2c^2.$$
It is a contraction.
