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

It can be done by differentiation but it is much easier to use $x^2 \geq 0$.
Assuming $a,b,c $ are positive
\begin{eqnarray*}
a(b-c)^2+b(c-a)^2+c(a-b)^2 \geq 0 \\
\end{eqnarray*}
Expand these out
\begin{eqnarray*}
ab(a+b)+bc(b+c)+ca(c+a) \geq 6abc \\
\end{eqnarray*}
Now divide by $abc$.
