A cyclic inequality: a²b + b²c + c²a + abc ≤ 4

Idea. Choose the variable in the middle and merge the other two.

Reduction. The expression is invariant under the cyclic shift $(a,b,c)\to(b,c,a)$, and one of the three numbers is between the other two, so we may assume $b$ is between $a$ and $c$. Then $(a-b)(b-c)\ge0$.

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

AM-GM. Hence the expression is at most $b(a+c)^2=4\cdot b\cdot\frac{a+c}2\cdot\frac{a+c}2\le4\left(\frac{b+(a+c)}3\right)^3=4$, since $b+(a+c)=3$.

Equality. We need $c(a-b)(b-c)=0$ and $b=\frac{a+c}2=1$. This gives $a=b=c=1$ or $c=0$, $b=1$, $a=2$ (and the cyclic shifts of the latter).

Lesson. For cyclic inequalities, "middle variable" arguments replace the symmetry you do not have: the sign of $(a-b)(b-c)$ gives the one extra inequality you need.