Find the maximum value of $xy^2z^3$ given that $x^2 + {y}^2 + {z}^2 = 1$, using AM-GM I've been struggling with this equation and how to find the maximum value it can take: 

Maximise $xy^2z^3$ given that $x^2+y^2+z^2 = 1$

The question is from the book Introduction to Inequalities - CJ Bradley. 
Any help would be much appreciated! 
Thanks 
 A: Since
$$
\frac16\left(6x^2\right)+\frac13\left(3y^2\right)+\frac12\left(2z^2\right)=1
$$
The AM-GM says that
$$
\left(6x^2\right)^{1/6}\left(3y^2\right)^{1/3}\left(2z^2\right)^{1/2}\le1
$$
which, upon raising to the third power, becomes
$$
xy^2z^3\le\frac1{12\sqrt3}
$$
The maximum is attained when $6x^2=3y^2=2z^2=1$.
A: To use AM=GM. rewrite the condition as
$$x^2+(1/2)y^2+(1/2)y^2+(1/3)z^2+(1/3)z^2+(1/3)z^2=1.$$
Applying AM=GM to the left side, we get that 
$$\frac{1}{6}=\frac{x^2+(1/2)y^2+(1/2)y^2+(1/3)z^2+(1/3)z^2+(1/3)z^2}{6}\ge \left((1/4)(1/27)x^2y^4z^6    \right)^{1/6}.$$
From this we can write down the answer. Note that we have equality when $|x|=\sqrt{1/2}\,|y|=\sqrt{1/3}\,|z|$.
A: Another non-AM/GM approach, also avoiding Lagrange multipliers, is to observe that the problem is equivalent to maximizing
$$f(x,z)=x(1-x^2-z^2)z^3$$
in the square region $0\le x,z\le1$.  
(Note:  The "Another" referred to an answer that has since been deleted.  André Nicolas's marvelous answer is, of course, the "right" one for the OP's purpose.)
