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 $$\frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(y+z)}}+\frac{z}{\sqrt{(x+z)(y+z)}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$$
 A: The desired inequality is written as
$$\frac{2x}{\sqrt{1 + x^2}}
+ \frac{2y}{\sqrt{1 + y^2}} + 
\frac{2z}{\sqrt{1 + z^2}}
+ 8 \cdot \frac{1 + x^2}{4x^2}
+ 8 \cdot \frac{1 + y^2}{4y^2}
+ 8 \cdot \frac{1 + z^2}{4z^2} \ge 27.$$
Using AM-GM, we have
\begin{align*}
 \mathrm{LHS}
 &\ge 27\,\sqrt[27]{\frac{2x}{\sqrt{1 + x^2}}
   \frac{2y}{\sqrt{1 + y^2}}
  \frac{2z}{\sqrt{1 + z^2}}
    \left(\frac{1 + x^2}{4x^2}\right)^8
   \left(\frac{1 + y^2}{4y^2}\right)^8
   \left(\frac{1 + z^2}{4z^2}\right)^8}\\
 &= 27\,\sqrt[27]{
 \left(\frac{\sqrt{(1 + x^2)(1 + y^2)(1 + z^2)}}{8xyz}\right)^{15}}\\
&= 27\,\sqrt[27]{
 \left(\frac{(x + y)(y + z)(z + x)}{8xyz}\right)^{15}}\\
&\ge 27\,\sqrt[27]{
 \left(\frac{2\sqrt{xy} \cdot 2\sqrt{yz} \cdot 2\sqrt{zx}}{8xyz}\right)^{15}}\\
&= 27
\end{align*}
where we have used
\begin{align*}
&(1 + x^2)(1 + y^2)(1 + z^2)\\
=\, & (xy + yz + zx + x^2)(xy + yz + zx + y^2)(xy + yz + zx + z^2)\\
=\, & (x + y)^2(y + z)^2(z + x)^2.
\end{align*}
We are done.
A: Let $x=\tan\frac{\alpha}{2},$ $y=\tan\frac{\beta}{2}$ and $z=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset\left(0,\frac{\pi}{2}\right).$
Thus, $\alpha+\beta+\gamma=\pi$ and since $$\left(\sin\frac{\alpha}{2}+\cot^2\frac{\alpha}{2}\right)''=\frac{2\cos{\alpha}+4-\sin^5\frac{\alpha}{2}}{4\sin^4\frac{\alpha}{2}}>0,$$ by Jensen we obtain:
$$\sum_{cyc}\left(\frac{x}{\sqrt{x^2+1}}+\frac{1}{x^2}\right)=\sum_{cyc}\left(\sin\frac{\alpha}{2}+\cot^2\frac{\alpha}{2}\right)\geq3\left(\sin\frac{\pi}{6}+\cot^2\frac{\pi}{6}\right)=\frac{21}{2}.$$
A: Another way.
Let $x^2+y^2+z^2=t(xy+xz+yz).$
Thus, $t\geq1$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}\left(\frac{x}{\sqrt{1+x^2}}+\frac{1}{x^2}\right)=\sum_{cyc}\frac{2x}{2\sqrt{(x+y)(x+z)}}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x^2y^2}{x^2y^2z^2}\geq$$
$$\geq\sum_{cyc}\frac{2x}{2x+y+z}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{xyz}=\sum_{cyc}\frac{2x^2}{2x^2+xy+xz}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{xyz}\geq$$
$$\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(2x^2+xy+xz)}+\frac{\sum\limits_{cyc}xy\sum\limits_{cyc}x}{\sqrt{\left(\frac{\sum\limits_{cyc}xy}{3}\right)^3}}=\frac{t+2}{t+1}+3\sqrt{3(t+2)}=$$
$$=\frac{t+2}{t+1}+3\sqrt{(1+2)(t+2)}\geq\frac{t+2}{t+1}+3(\sqrt{t}+2)\geq$$
$$\geq\frac{t+2}{t+1}+3\left(\frac{2t}{t+1}+2\right)=\frac{5t}{t+1}+8\geq\frac{5t}{t+t}+8=\frac{21}{2}.$$
A: 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 .
