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. 

This is the same as
$$\left(\frac a6+\frac b3+\frac c2\right)(bc+2ca+3ab)\ge 6abc$$
which follows from the Cauchy-Schwarz inequality applied to the vectors
$$\left(\sqrt{\frac a6},\sqrt{\frac b3},\sqrt{\frac c2}\right)
\qquad\text{and}\qquad(\sqrt{bc},\sqrt{2ca},\sqrt{3ab}).$$
