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?

Your solution is wrong because in your solution equality occurs when $a=b=1$ which is not possible because $x=y=z$ doesn't satisfy the constraints. Instead, you should use method of lagrange multipliers. Let $m = \frac{y}{x}$ and $n = \frac{z}{x}$. Then, lagrangian
$$L = m+n-\lambda(3(1+m+n)-mn)$$
(the constraint is obtained by dividing the given equation by $x^2$ on both sides)Equations formed after differentiating $L$ with respect to $m, n$ and $\lambda$ are $$1 =\lambda(3-n)$$
$$1 = \lambda(3-m)$$
$$3(1+m+n)=mn$$
From first two equations, $m=n$, using it in third equation, we get,
$$m^2-6m-3=0$$
$$\implies m = 3+2\sqrt{3}$$
Required minimum value is $6+4\sqrt{3}$.
