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?
On
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 $6+4\sqrt{3}$.
On
Use Lagrange multipliers. Call the function you want to minimize $f$ and the constraint $g$. From the condition on the partial derivatives on $y$ and $z$ of $f - \Lambda g$ you can deduce that $y=z$. Finding $x$ with respect to $y$ gives you $x=\frac{\sqrt{12}-3}{3}y$, and this leads to the solution. You'd have to check this is actually a minimum (in principle this is only a critical point), but you should be able to verify that by yourself.
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.