prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$

In this case you want to have quadratic terms in the RHS, so use Cauchy in the following form: $$\left(\sum_{cyc} \frac{a^2}b \right) \left(\sum_{cyc} a^2b \right) \ge \left(a^2+b^2+c^2\right)^2 \tag{1}$$

Then it is enough to show that $$a^2+b^2+c^2 \ge 3\sum_{cyc} a^2b \tag{2}$$ Homogenising, $$\iff \left( a^2+b^2+c^2\right) (a+b+c) \ge 3\sum_{cyc} a^2b $$ $$\iff \sum_{cyc} a^3+ \sum_{cyc} ab^2 \ge 2\sum_{cyc} a^2b $$

which follows from AM-GM as $a^3+ab^2 \ge 2a^2b$.

lemma: $$\sum_{cyc}\dfrac{a^2}{b}\ge\dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}$$ Proof: $$\Longleftrightarrow (ab+bc+ac)(\sum_{cyc}\dfrac{a^2}{b})\ge (a+b+c)(a^2+b^2+c^2)$$ $$\Longleftrightarrow a^3+b^3+c^3+a^2c+c^2b+b^2a+\sum_{cyc}\dfrac{a^3c}{b}\ge a^3+b^3+c^3+\sum_{sym }a^2b$$ $$\Longleftrightarrow\dfrac{a^3c}{b}+\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge ac^2+cb^2+ba^2$$ By AM-GM,we have $$\dfrac{a^3c}{b}+\dfrac{b^3a}{c}\ge 2a^2b$$ $$\dfrac{b^3a}{c}+\dfrac{c^3b}{a}\ge 2b^2c$$ $$\dfrac{a^3c}{b}+\dfrac{c^3b}{a}\ge 2c^2a$$ if we prove this $$\dfrac{a+b+c}{ab+bc+ac}=\dfrac{(a+b+c)^2}{ab+bc+ac}\ge 3$$