Want less brutish proof: if $a+b+c=3abc$ then $\frac1a+\frac1b+\frac1c\geq 3$

Question

In the course of showing for positive $a,b,c\,$ that $a+b+c=3abc$ implies $$\frac1{2a+b} + \frac1{2b+c} + \frac1{2c+a} \geq 1$$ I needed to show that (given that same constraint)

$$ \frac1a+\frac1b+\frac1c\geq 3$$

I have a proof which is human-readable but less elegant than I would like: $$ 3\left( ab-\frac12(a+b)\right)^2 +\frac14(a-b)^2 \geq 0 $$ rearranges to $$ -3ab(a+b) + (a+b)^2 + \left[ 3a^2b^2-ab\right] \geq 0 $$ and dividing by the positie quantity $ab(a+b)$ this becomes $$ \frac1a + \frac1b + \frac{3ab-1}{a+b} \geq 3 $$ and the third term on the right is, by the constraint, equal to $\frac1c$.

What I dislike is that to find that first expression, I used a Buffalo-like technique: I solved the constraint for $c$ in terms of $a$ and $b$, substituted in the inequality, multiplied through by the product of denominators $ab(a+b)$, found the first term in the sum of squares by demanding that the $a^2b^2$ and $(a^2b+b^2a)$ terms be covered.

By some lucky chance the remaining terms were a recognizable square (although the Buffalo Way could also handle remaining terms with all positive coefficients).

It seems to me that there should be a better way of showing this, perhaps using Jensen's inequality, valid for convex upward functions $f(t)$: $$ f\left(\frac{\sum w_ix_i}{\sum w_i} \right)\leq \frac{\sum w_if(x_i)}{\sum w_i} $$ The usual trick is to use some simple $f(t)$ like $t^{-2}$ and choose the weights and the $x_I$ to make the numerator on the right hand side come out to the business end of the inequality (and the other sums expressions that can be handled using the given constraint). But I can't find the way to do that here.

Can anybody find a proof that is a bit less ad hoc in its starting point?

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ADDED LATER

After seeing the excellent proofs offered by Arnaldo and Michael, I realized that although this was nice to prove, it in fact does not help me prove that $$\frac1{2a+b} + \frac1{2b+c} + \frac1{2c+a} \geq 1$$

Oh well...

Answer 1

I will use that

$$a+b+c=3abc \Leftrightarrow \frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3$$

By AM-GM $$\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\frac{4}{ab}\\ \left(\frac{1}{a}+\frac{1}{c}\right)^2\ge\frac{4}{ac}\\ \left(\frac{1}{b}+\frac{1}{c}\right)^2\ge\frac{4}{bc}$$

so

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2\ge 4\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=12\quad (1)$$

but $x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)$, so

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+9\right)\quad (2)$$

and also,

$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{ab}+\frac{1}{bc}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\quad (3)$$

Now use $(3)$ in $(2)$ and get:

$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6 \quad (4)$$

Now put $(4)$ in $(1)$:

$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\ge 12\to \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3$$

Answer 2

We need to prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq3\sqrt{\frac{a+b+c}{3abc}}$$ or $$ab+ac+bc\geq\sqrt{3abc(a+b+c)}$$ or $$\sum_{cyc}(a^2b^2-a^2bc)\geq0$$ or $$\sum_{cyc}(a^2c^2+b^2c^2-2c^2ab)\geq0$$ or $$\sum_{cyc}c^2(a-b)^2\geq0.$$ Done!