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.

Consider you that you look for the minimum of function $$\Phi=\frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \tag 1$$ Compute the partial derivatives 
$$\frac{\partial\Phi}{da}=-\frac{b+c}{a^2}+\frac{1}{b}+\frac{1}{c}\tag 2$$
$$\frac{\partial\Phi}{db}=-\frac{a+c}{b^2}+\frac{1}{a}+\frac{1}{c}\tag 3$$
$$\frac{\partial\Phi}{dc}=-\frac{a+b}{c^2}+\frac{1}{a}+\frac{1}{b}\tag 4$$ and say that all of them are equal to $0$.
From $(2)$ solve the quadratic for $b$. The two roots are
$$b_1=\frac {a^2} c\qquad \text{and}\qquad b_2=-c\implies b=\frac {a^2} c$$ since $a,b,c$ are positive.
Plug $b=\frac {a^2} c$ in $(3)$ to get $$\frac{(a+c) \left(a^3-c^3\right)}{a^4 c}=0\implies c=a\implies b=c=a$$
Plug in $(4)$ to get $0=0$.
Replace in $(1)$ and get $6$.
