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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.

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Apply $AM \ge HM$ to the $6$-tuple $(a,b,b,c,c,c)$, one get

$$LHS = \frac16(a+2b+3c) \ge \frac{1}{\frac16(\frac{1}{a}+\frac{2}{b}+\frac{3}{c})} = \frac{6abc}{bc+2ac+3ab} = RHS $$

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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.

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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 $$\left(\sqrt{\frac a6},\sqrt{\frac b3},\sqrt{\frac c2}\right) \qquad\text{and}\qquad(\sqrt{bc},\sqrt{2ca},\sqrt{3ab}).$$

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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.

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