Mixing variables: a²b + b²c + c²a ≤ 4/27

Reduction. The expression is invariant under the rotation $a\to b\to c\to a$. Among three numbers, one is between the other two; by rotating, we may assume $b$ lies between $a$ and $c$, so $(a-b)(b-c)\ge0$.

Key identity. Expanding,
$$b(a+c)^2-\left(a^2b+b^2c+c^2a+abc\right)=c(a-b)(b-c)\ge0.$$

Conclusion. Since $abc\ge0$,
$$a^2b+b^2c+c^2a\le b(a+c)^2=4\cdot b\cdot\frac{a+c}2\cdot\frac{a+c}2\le4\left(\frac{b+\frac{a+c}2+\frac{a+c}2}3\right)^3=\frac4{27}.$$

Equality. We need $abc=0$, $c(a-b)(b-c)=0$ and $b=\frac{a+c}2$ simultaneously. This gives $(a,b,c)=\left(\frac23,\frac13,0\right)$ and its cyclic rotations.

Remark. The maximum is *not* at $a=b=c$ (where the value is $\frac19=\frac3{27}$): it is on the boundary.

Lesson. Cyclic inequality with an equality case on the boundary: order the variables (carefully, only using rotations) and push the expression to a two-variable one.