<s>[INST] Find minimum value of $P = \frac{y + z}{x}$ 
Let $x,y,z > 0$ such that $3x(x+y+z)=yz$ . Find the minimum value of $$P=\frac{y+z}{x}$$


$$3x(x+y+z)=yz\Leftrightarrow 3(x+y)(x+z)=4yz$$
Or $$3(\frac{x}{y}+1)(\frac{x}{z}+1)=4$$
Let $a=x/y, b=x/z ( a,b>0)$
We have: $$4=3(a+1)(b+1)\ge3 \cdot 2 \sqrt  a \cdot 2 \sqrt  b=12 \sqrt {ab} \rightarrow \sqrt {ab}\le \frac 1 3$$
And $$P=\frac{y+z}{x}=\frac{1}{a}+\frac{1}{b}\ge \frac{2}{\sqrt {ab}}\ge 6$$
But by WA $Min_P=6+4\sqrt 3$. Is my solution is wrong?
 [/INST] Let $y+z=2u$ and $yz=v^2$, where $v>0$.
Thus, since by AM-GM $$\frac{y+z}{2}\geq\sqrt{yz},$$ we obtain $u\geq v$, 
$$3x^2+6ux-v^2=0$$ or $$x=-u+\sqrt{u^2+\frac{v^2}{3}}$$ and
$$\frac{y+z}{x}=\frac{2u\left(\sqrt{u^2+\frac{v^2}{3}}+u\right)}{\frac{v^2}{3}}\geq\frac{2v\left(\sqrt{v^2+\frac{v^2}{3}}+v\right)}{\frac{v^2}{3}}=6+4\sqrt3.$$ 
The equality occurs for $u=v$ or $y=z$, which says that we got a minimal value.
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