Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$ Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$  
 A: By AM-HM or C-S we have
$$\frac{\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}}{3} \geq \frac{3}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
Therefore 
$$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}} \geq \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$
To prove your claim, you need to show
$$\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \leq \sqrt{10}$$
By C-S
$$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}  \right)^2 \leq 3 (1+xy+1+yz+1+zx )=9+3(xy+xz+yz)$$
Finally, again by C-S we have
$$xy+xz+yz \leq x^2+y^2+z^2 \Rightarrow 3\\(xy+xz+yz) \leq x^2+y^2+z^2+2(xy+xz+yz) =(x+y+z)^2=1$$
Combining the last two inequalities we get
$$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}  \right)^2 \leq 10$$
A: By the inequality between harmonic and arithmetic means we have that 
$$
\frac{1}{\sqrt{1+xy}}+\frac{1}{\sqrt{1+yz}}+\frac{1}{\sqrt{1+xz}}\ge \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}}.
$$
Now if we prove that 
$$
\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{10}
$$
we are done. but by Cauchy-Schwarz we have 
$$
\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{3}(3+xy+yz+zx)^{\frac{1}{2}},
$$
so we just need to prove that 
$$
xy+yz+zx\le\frac{1}{3},
$$
but it is the same to prove that 
$$
x^2+y^2+z^2\ge\frac{1}{3}.
$$
The last inequality can be easily proved using Cauchy-Schwarz.
A: Here is another approach.
Cauchy-Schwarz says
$$
\begin{align}
(x+y+z)^2
&\le(x^2+y^2+z^2)(1^2+1^2+1^2)\\
&=3(x^2+y^2+z^2)\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
xy+yz+zx
&=\tfrac12\left[(x+y+z)^2-\left(x^2+y^2+z^2\right)\right]\\
&\le\tfrac12\left[(x+y+z)^2-\tfrac13(x+y+z)^2\right]\\
&=\tfrac13(x+y+z)^2\tag{2}
\end{align}
$$
Then
$$
\begin{align}
\frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}
&\ge\frac3{\sqrt[\large6]{(1+xy)(1+yz)(1+zx)}}\tag{3}\\[3pt]
&\ge\frac3{\sqrt{\frac13\left[(1+xy)+(1+yz)+(1+zx)\right]}}\tag{4}\\
&=\frac9{\sqrt{9+3(xy+yz+zx)}}\tag{5}\\[3pt]
&\ge\frac9{\sqrt{10}}\tag{6}
\end{align}
$$
Explanation:
$(3)$: AM-GM
$(4)$: AM-GM in the denominator inside the square root
$(5)$: regrouping
$(6)$: apply $(2)$
