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.
 A: 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 
$$
t+\frac{1}{t}\ge 2,\quad t>0.
$$
It gives the result.
A: 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 $6$.
A: 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 $abc$.
A: 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).
A: By AM-GM for positive variables:
$$\sum_{cyc}\frac{b+c}{a}\geq2\sum_{cyc}\frac{\sqrt{bc}}{a}\geq6\sqrt[3]{\prod_{cyc}\frac{\sqrt{bc}}{a}}=6.$$
AM-GM it's the Jensen inequality for $f(x)=\ln{x}$ because $f''(x)=-\frac{1}{x^2}<0$.
