<s>[INST] 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? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$  
 [/INST] 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 $(2)$
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