Prove that: $\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$ Given 3 positive real numbers $x, y, z$ satisfies $xy+yz+xz=1$. Prove that: $$\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq\frac{21}{2}$$
(Only use AM-GM, Cauchy-Schwarz inequalities)
My progress:
Till now, I have not made much of a progress besides finding out that $x^2+y^2+z^2\geq1$ and $xyz\leq\frac{\sqrt{3}}{9}$ and turn the LHS into $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$

Hint :
If we have :
$$f\left(x\right)=\frac{x}{\sqrt{x^{2}+1}}+\frac{1}{x^{2}}$$
And define :
$$h\left(x,y\right)=f\left(\frac{1-xy}{x+y}\right)-\frac{21}{2}xy\left(1+C\frac{\frac{2}{21}-\frac{2}{63xy}}{x^{2}+y^{2}+\left(\frac{1-xy}{x+y}\right)^{2}}\right)$$
Then with the constraint $0<xy\le1$  and $6<C<7$ a judicious constant try to show:
$$h\left(x,y\right)\ge0$$
Now use it three times .
For $0<x\leq 1$ and a end see Wolfram Alpha transformation
I recognize it's very ugly but all the good way are complete see other answer .
