if $x+y+z=1$ then find the maximum of the form 
Let $x>0$, $y>0$ and $z>0$ such that $x+y+z=1$. Find the maximal value of
  $$x\sqrt{y}+y\sqrt{z}$$

I think $x=y=\dfrac{4}{9},z=\dfrac{1}{9}$ then the maximum $\dfrac{4}{9}$,but how to use AM-GM prove it?
 A: By AM-GM
$$x\sqrt{y}+y\sqrt{z}=\sqrt{y}\left(x+2\sqrt{\frac{y}{4}\cdot{z}}\right)\leq\sqrt{y}\left(x+\frac{y}{4}+z\right)=$$
$$=\sqrt{y}\left(1-y+\frac{y}{4}\right)=\frac{4}{9}-\frac{1}{36}\left(\left(3\sqrt{y}\right)^3+16-36\sqrt{y}\right)\leq$$
$$\leq\frac{4}{9}-\frac{1}{36}\left(3\sqrt[3]{\left(3\sqrt{y}\right)^3\cdot8^2}-36\sqrt{y}\right)=\frac{4}{9}.$$
The equality occurs for $x=y=\frac{4}{9}$ and $z=\frac{1}{9}$, which says that $\frac{4}{9}$ is the maximal value.
Done!
A: Without AM-GM, the Lagrange multiplier method will do:
$$L=x\sqrt{y}+y\sqrt{z}+t(1-x-y-z).$$
$$\begin{cases} L_x=\sqrt{y}-t=0 \\ L_y=\frac{x}{2\sqrt{y}}+\sqrt{z}-t=0 \\ L_z=\frac{y}{2\sqrt{z}}-t=0 \\ L_t=1-(x+y+z)=0 \end{cases} \Rightarrow x=y=\frac{4}{9}, z=\frac{1}{9}.$$
Note: To solve the system of equations, square $(1)$ divide it by $(3)$ to find $t=2\sqrt{z}$. Then substitute this into $(1)$ to find $y=4z$. Then substitute these into $(2)$ to find $x=4z$. And finally substitute all into $(4)$ to find $z=\frac{1}{9}$, hence the rest.
