Weighted AM-GM: how to choose the splitting

1. Apply AM-GM to the five numbers $\frac x2,\frac x2,\frac y3,\frac y3,\frac y3$, whose sum is $x+y=5$:
$$\left(\frac x2\right)^2\left(\frac y3\right)^3\le\left(\frac{x+y}5\right)^5=1.$$
Hence $x^2y^3\le4\cdot27=108$. Equality iff all five numbers are equal: $\frac x2=\frac y3$, i.e. $x=2$, $y=3$. The maximum is $108$.

2. AM-GM on the five numbers $a,a,b,b,b$ gives
$$\frac{2a+3b}5\ge\sqrt[5]{a^2b^3}\iff2a+3b\ge5a^{2/5}b^{3/5}.$$
In general, for $\lambda\in(0,1)$: $\lambda a+(1-\lambda)b\ge a^\lambda b^{1-\lambda}$.

Remark. Compare with calculus: maximizing $x^2(5-x)^3$ by differentiation gives the same point.

Lesson. To maximize a product under a sum constraint, split the variables into equal parts proportional to their exponents.