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

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?

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...

• Theorem: There must be an infinite number of contest-math problems containing the letters $a,b,c$ which are invariant under (at least) cyclic permutations. Proof: Every time I look, I find another one. And on a more somber note: These problems are brutal, boring, lack structure, teach nothing, mean nothing, and prove nothing. They must the ignoblest problems ever. – uniquesolution Mar 28 '17 at 20:47
• @uniquesolution ... For goodness sake ... Don't let Michael Rozenberg hear you say such a thing ... some of us really enjoy these seemingly valuless truths :-) – Donald Splutterwit Mar 28 '17 at 21:02
• These problems are an excellent tools for assessing the degree of degradation of one's mental facilities. Way back when, in my school math contest and Putnam exam days, this sort of problem was almost a "gimme" for me. As I am getting older, the problems are getting to be quite difficult. To me, these are more fun than doing crosswords or quote-acrostics to measure how rapidly I grow senile. – Mark Fischler Mar 28 '17 at 21:14
• @uniquesolution... Your first premise is not quite accurate. Some of the problems use $x,y,z$. – Mark Fischler Mar 28 '17 at 21:39
• @MarkFischler Clearly, those $\sum_{cyc}$ problems do have their lasting impact: how on earth can the existence of a few $xyz$ problems contradict the fact that every time I look, I see an $abc$ one? – uniquesolution Mar 29 '17 at 6:22

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 $(1)$:

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

• This can be made more concise -- equation (4) follows immediately from the constraint, which can be written as $\sum\frac1{ab} = 3$. Equation (1) then scores the goal. Still, you get the point for a solution which is way less ad hoc than mine. – Mark Fischler Mar 28 '17 at 21:37
• @MarkFischler: you are right! – Arnaldo Mar 28 '17 at 21:39

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!