Geometry Problem from China high school mathematics tournament

Let $AB$ be a chord in the circle $A_{1}$; $AE=EF=FB$; Point $X$ is on the circle, $PE$ meet with the circle at point $Y$, $PF$ meet with the circle at point $Z$. Show that $AY\cdot BZ=EF\cdot YZ$.

Proof:  Apply the law of sine, we have

\[\frac{\sin{\angle AZY}}{\sin{\angle{YAZ}}}=\frac{AY}{YZ}=\frac{\sin{\angle AXY}}{\sin{\angle YXZ}}\]

\[=\frac{XF}{XA}=\frac{\sin\angle XAE}{\sin\angle XFA}=\frac{BF}{BZ}=\frac{EF}{BZ}\]

Comment: This problem is not kind of hard. It can be proved with pure geometry method easily too. Just consider the rotation of $\Delta AXZ$  

prove the minimum value by completing square

let $x,y$be positive poof that \[x^3+y^3+xy^2+x^2y-2x^2-2y^2-2xy+2\geqslant0\]

Poof: it can be written as \[\frac{1}{4}(x-1)^2(3x+y)+\frac{1}{4}(y-1)^2(3y+x)+\frac{1}{4}(x+y+2)(x-2+y)^2\], done