<s>[INST] Prove this Hard inequality to CS or AM-GM? 
Let $x,y,z\ge 0$ such $x+y+z=1$, show that
  $$\sqrt{\dfrac{x(x+1)}{1-x}}+\sqrt{\dfrac{y(y+1)}{1-y}}+\sqrt{\dfrac{z(z+1)}{1-z}}\ge\sqrt{6}$$

This inequality is creat by wangyongxi,and  when $x=y=z=\dfrac{1}{3}$ or $x=y=\dfrac{1}{2},z=0$ is $=$.
and we found this $f(x)=\sqrt{\dfrac{x(x+1)}{1-x}},0\le x\le 1$,not Convex function,because we get
$$f''(x)=-\dfrac{x^4-4x^3-6x^2-4x+1}{4(x-1)^4\left(-\dfrac{x(x+1)}{x-1}\right)^{3/2}}$$
so when $x<0.2$then $f''(x)<0$ and other $f''(x)>0$
Now I wonder to look such $AM-GM$ or Cauchy inequality to solve it? Thanks
 [/INST] Write a tangent line at $T(\sqrt{2}-1,1)$ for $f(x)= \sqrt{\dfrac{x(x+1)}{1-x}} $ 
$$ y= (\sqrt{2}+1)x$$
and see that it is entirely under the graph of $f$ on $[0,1]$: 
$$\sqrt{\dfrac{x(x+1)}{1-x}}\geq (\sqrt{2}+1)x$$
So $$f(x)+f(y)+f(z)\geq \sqrt{2}+1$$
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