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.
 A: 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.
A: $a=\sqrt{x+1}>1,b=\sqrt{y+1}>1, c=\sqrt{z+1}>1$,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.
A: 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.
A: 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.
A: The problem can be restated as follows. Find the maximum of the function
$$
f(x,y,z)=\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}
$$
constrained to $xyz-8=0$. You can use a Lagrange multiplier and proceed. 
