Show $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} \ge 3$ for $a,b,c > 0$. Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it. 
My workings:
I got a common denominator and expressed it as:
$$\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3$$
$$ab^2 + bc^2 + ca^2 \ge 3abc$$
This is where I'm stuck. I feel as if I keep encountering this same result in various different problems and always am stumped on how to show it is true. 
I tried to look at similar problems and it seems AM-GM inequality is the usual line of attack, so using that I reasoned, $$a^2 + b^2 + c^2 \ge 3abc$$, and now I need to establish $$ca^2 + ab^2 + bc^2 \ge a^2 + b^2 + c^2$$
However, this doesn't seem like it'd be true for general positive $a,b,c$.
 A: 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 $a=2, b=2, c=1$.
A: 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 $(a,b,c) \in \mathbb{R}^3$.
A: 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
A: Note that if $x=\frac bc, y=\frac ca, z=\frac ab$ we have $xyz=1$ and we apply AM-GM to $x,y,z$ we obtain $$\frac {x+y+z}3\ge (xyz)^{\frac 13}=1$$
A: By Rearrangement inequality
$$\frac{b}{c} + \frac{c}{a} + \frac{a}{b} \ge \frac{a}{a} + \frac{b}{b} + \frac{c}{c}=1+1+1=3$$
