Inequality question involving maximum value and three variables

Question

If $x, y, z > 0$ and $x + y + z = 1$, prove that

(a) $x^2+y^2+z^2 \geq \frac{1}{3}$

(b) $x^2yz \leq \frac{1}{64}$

For the first part, since

$\begin{align} x+y+z=1 &\geq 3\cdot \sqrt[3]{xyz} \\ \Rightarrow \frac{1}{9} &\geq \sqrt[3]{x^2y^2z^2} \end{align}$

and thus

$\begin{align} x^2+y^2+z^2 &\geq 3 \cdot \sqrt[3]{x^2y^2z^2} \\ \therefore x^2+y^2+z^2 &\geq \frac{1}{3} \blacksquare. \end{align}$

I dont understand how i should proceed for the second part tho. Here is my attempt at it with weighted AM GM.

$\begin{align} 2x+y+z &\geq 4 \cdot \sqrt[4]{x^2yz} \\ \frac{1+x}{4} &\geq \sqrt[4]{x^2yz} \\ \frac{x^4+4x^3+6x^2+4x+1}{256} &\geq x^2yz \\ \end{align}$

But i don't get how i should proceed from here. Please check the solutions and help with the second part. Thanks

Answer 1

For (b) apply AM/GM to $x$, $x$, $2y$ and $2z$. We get $$\frac{2(x+y+z)}{4}\ge\sqrt[4]{4x^2yz}$$ etc.

Answer 2

$$\dfrac{2\times\frac{x}{2}+y+z}{4}\geq \sqrt[4]{\frac{x^2}{4}yz}=\frac{\sqrt[4]{x^2yz}}{\sqrt{2}}\implies x^2yz\leq \frac{4}{4^4}=\frac{1}{64}$$