Prove that $a^2b + b^2c + c^2a \ge ab + bc+ ac$ for positive real numbers $a,b,c$ such that $a+b+c=3$.
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The theorem, as stated, is false, as the following counterexample shows.
First note that both sides of the inequality and the restriction $a + b + c = 3$ are continuous functions of $a$, $b$, and $c$, so we can consider non-negative values rather than positive. Then if $a = 0$ we have $$ b^2 c \geq bc $$ for any $b + c = 3$. But this is false if $c > 0$ and $1 > b > 0$, for example $c = 2 + \frac12$ and $b = \frac12$. Note that you can translate this to a counterexample for the original problem by setting $a = \epsilon$ and $c = 2 + \frac12 - \epsilon$ for $\epsilon > 0$ sufficiently small.
There are a couple of ways to construct similar (correct) inequalities so there's probably a typo in your problem source. In general you can use the AM-GM inequality to solve problems like this one, or the Muirhead inequality. You might try this paper for an overview.
You can use Lagrange multipliers. Define $f(a,b,c)=a^2b+b^2c+c^2a-ab-bc-ca$. Since the triangle $a+b+c=3$ with $a,b,c \ge 0$ is compact, f has a global minimum on this domain. If this minimum lies on the boundary, WLOG at $a=0$, it's easy to see that $f \ge 0$ on this line. So if $f$ obtains its minimum at a point $(a,b,c)$ in the interior of the triangle, the method of Lagrange multipliers tells us that there must be a $lambda \in \Bbb R$ such that $\lambda (a,b,c)=\nabla f(a,b,c)=(2ab+c^2-b-c,2bc +a^2-c-a,2ca +b^2 -a -b)$. Solving this system of equations for $a,b,c$ (remember that you have $a+b+c=3$ in addition, so that you have 4 equations and 4 variables) and checking that $f \ge 0$ on all these solutions completes the proof.