<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 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.
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