<s>[INST] 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.
 [/INST] $a=\sqrt{x+1}>1,b=\sqrt{y+1}>1, c=\sqrt{z+1}>1$,for $a,b,c$,at least two of them $>2$ or $\le 2$,WOLG, assume $a,b$ either both $>2$ or both $\le2$, $z=\dfrac{8}{xy}=\dfrac{8}{(a^2-1)(b^2-1)}$, 
edit:
case I: 
if $a>2$ and $b>2$  
it is trivial $\dfrac{1}{a}+\dfrac{1}{b} <1 ,\dfrac{1}{c}<1$
so it is true in this case.
case II: when $a \le 2,b \le 2$, 
now we need to prove:
$\dfrac{1}{a}+\dfrac{1}{b}+\sqrt{\dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}}<2 \iff \dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}< \left(2-\dfrac{1}{a}-\dfrac{1}{b} \right)^2 \iff \dfrac{(2ab-a-b)^2}{a^2b^2} > \dfrac{(a^2-1)(b^2-1)}{((a^2-1)(b^2-1)+8)}$ 
note: $(a^2-1)(b^2-1)+8>a^2b^2 \iff 9>a^2+b^2 \iff (a^2\le 4) \cap (b^2 \le 4) $
now we need to prove  $(2ab-a-b)^2>(a^2-1)(b^2-1)$
$(2ab-a-b)^2= (a(b-1)+b(a-1))^2 \ge 4ab(a-1)(b-1) \iff 4ab >(a+1)(b+1) \iff (2a>a+1 )\cap (2b>b+1) \iff (a>1) \cap (b>1)$ 
it is true. 
both cases are true. QED
Indeed, when $xyz=2$, this inequality is also true.
</s>