a²/b + b²/c + c²/a ≥ a + b + c

Titu with $(x_i)=(a,b,c)$ and $(y_i)=(b,c,a)$.
$$\frac{a^2}b+\frac{b^2}c+\frac{c^2}a\ge\frac{(a+b+c)^2}{b+c+a}=a+b+c.$$

Second proof (tangent trick). For positive $x,y$: $\frac{x^2}y\ge2x-y$, because $\frac{x^2}y-2x+y=\frac{(x-y)^2}y\ge0$. Hence
$$\sum\frac{a^2}b\ge(2a-b)+(2b-c)+(2c-a)=a+b+c.$$

Equality iff $a=b=c$.

Remark. The second proof shows the strength of the inequality $\frac{x^2}y\ge2x-y$: it is the tangent line at $y=x$ of the convex function $y\mapsto\frac{x^2}y$.

Lesson. Know the two faces of the same tool: Titu (global) and $\frac{x^2}y\ge2x-y$ (term by term).