Proving $\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca}$ for positive $a$, $b$, $c$ I'm at the end of an inequality proof that started out complex and I was able to simplify it to:

$$\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c \geq \frac{6abc}{3ab+bc+2ca} \quad\text{where}\quad a, b, c > 0$$

I'm able to plug in very small values close to 0 and the inequality holds, but I'm having trouble finding a way to prove it and how to start off this problem. 

A straightforward proof can be posed as follows, the inequality can be written as such 
$$
(3ab+bc+2ca)\left(\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c\right) \ge 6abc
$$
which is equivalent to 
$$
3a^2b + 6ab^2 + 2ca^2 + 6c^2a + 2b^2c + 3b^2c \ge 22 abc.
$$
Now using that fact that $P(2, 1, 0) \ge P(1, 1, 1)$ (known as Muirhead's inequality) and $$P(\alpha, \beta, \gamma) := \sum_{\text{sym}}a^{\alpha}b^{\beta}c^{\gamma},$$ e.g. $P(1, 1, 1) = 6abc$, $P(2, 1, 0) = a^2b + ab^2 + bc^2 + b^2c + ac^2 + a^2c$. Thus, we have 
$$
\underbrace{a^2b + b^2c}_{\ge 2abc} + \underbrace{4ab^2 + 4c^2a}_{\ge 8abc}\ge 10 abc,
$$
as required.
