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. 

Since $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}=1,$ by AM-GM we obtain:
$$\left(\frac{1}{6}a+\frac{1}{3}b+\frac{1}{2}c\right)\left(\frac{1}{2}ab+\frac{1}{6}bc+\frac{1}{3}ca\right)\geq$$
$$\geq a^{\frac{1}{6}}b^{\frac{1}{3}}c^{\frac{1}{2}}(ab)^{\frac{1}{2}}(bc)^{\frac{1}{6}}(ca)^{\frac{1}{3}}=abc,$$ which gives your inequality.
Also, by C-S
$$\left(\frac{a}{6}+\frac{b}{3}+\frac{c}{2}\right)\left(\frac{1}{6a}+\frac{1}{3b}+\frac{1}{2c}\right)\geq\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)^2=1,$$ which gives your inequality again.
