Show $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} \ge 3$ for $a,b,c > 0$.
Apply AM-GM inequality:
$$ab^2+bc^2+ca^2\geq 3 \sqrt[3]{ab^2 \cdot bc^2 \cdot ca^2} = 3abc$$
Done! :)
It is not true that AM-GM inequality gives $a^2+b^2+c^2\geq 3abc$. This inequality is false: Let $a=2, b=2, c=1$.
It's not too hard to prove this using calculus. If $$\mathrm{f}(a,b,c) = \frac{b}{c}+\frac{c}{a}+\frac{a}{b}$$ then the gradiant vector is given by
$$\nabla\mathrm{f} = \left(\frac{a^2-bc}{a^2b}, \frac{b^2-ac}{b^2c}, \frac{c^2-ab}{ac^2}\right)$$
Solving $\nabla{\mathrm{f}}={\bf 0}$ gives only one real solution, namely $a=b=c$.
Calculating the Hessian Matrix and putting $a=b=c$ gives
$$H = \frac{1}{a^2}\left(\begin{array}{ccc} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2\end{array}\right)$$
The Hessian Matrix is singular, i.e. $\det H = 0$, and its kernel is spanned by $(1,1,1)^{\top}$.
This makes perfect sense. The line $(t,t,t)$ is sent to a "valley" on the graph $z=\mathrm{f}(a,b,c)$. The image $\mathrm{f}(t,t,t)$ is a straight line of minimum points. Direct calculation shows that $\mathrm{f}(t,t,t)=3$ while, for example $\mathrm{f}(1,2,1) = \frac{7}{2} > 3$. It follows that $\mathrm{f}(a,b,c) \ge 3$ for all $(a,b,c) \in \mathbb{R}^3$.
By AM-GM inequality $\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}{3}\geq\sqrt[3]{\frac{a}{b} \cdot \frac{b}{c}\cdot\frac{c}{a}}$
Therefore $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3$
Q.E.D