Can the inequality $\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6$ be proved with differentiation? $$\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6,\quad \text{with}\quad a,b,c &gt;  0$$

I could do it with letting $x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$, but I wonder if it is solvable somehow with differentiation.
Yes, you can do it by differentiation too. In your terms, the LHS becomes
$$
x+\frac{1}{x}+y+\frac{1}{y}+z+\frac{1}{z}.
$$
By differentiation, you can show that 
$$
t+\frac{1}{t}\ge 2,\quad t&gt;0.
$$
It gives the result.
This is a standard example: it's equivalent to
$$(a+b+c)\left(\frac1a+\frac1b+\frac1c\right)\ge9$$
which is AM/HM or Cauchy-Schwarz (provided all $a$, $b$, $c$ are non-negative).