Context. I saw a problem on facebook.It's
Let positive real numbers $x,y,z$ such that $x+y+z=x^2+y^2+z^2$. Prove that$$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{3x^3+6xyz}+\sqrt{3y^3+6xyz}+\sqrt{3z^3+6xyz}.$$ Source: Phan Ngoc Chau.
Here is my attempt.
Since equality occurs at $x=y=z=1,$ I used AM-GM $2\sqrt{3x^3+6xyz}=2\sqrt{3x(x^2+2yz)}\le 3x+x^2+2yz.$
Similarly, we need to prove $$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge\frac{(x+y+z)^2+3(x+y+z)}{2}.$$ Or$$\frac{xy+yz+zx+6xyz}{\sqrt{xyz}}\ge\frac{(x+y+z)^2+3(x+y+z)}{2}.$$ Let $a+b+c=p; ab+bc+ca=q; abc=r.$ By the hypothesis $p^2=2q+p\implies q=\dfrac{p^2-p}{2}$ and we'll prove $$\frac{p^2-p+12r}{\sqrt{r}}\ge p^2+3p.$$I was stuck here.
Can you help me? Thank you.
Update.
As Calvin Lin point out, my approach is not correct. Hope there is a better idea.
The C-S below doesn't help.$$\sqrt{\frac{xy}{z}}+\sqrt{\frac{yz}{x}}+\sqrt{\frac{zx}{y}}+6\sqrt{xyz}\ge \sqrt{3}\cdot\sqrt{3(x^3+y^3+z^3)+18xyz}.$$