<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] 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
$$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}  \right)^2 \leq 10$$
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