Prove the sequence of three real numbers

Question

If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then prove that $a,b,c$ are terms in $G.P$

My work:

I assumed that they are in $G.P$ and so assumed $b=ak$ and $c=ak^2$ for some arbitrary $k$. After that I expanded both sides of the equality and got the same results so that means the equality is true. But here's the twist!

Suppose the question was

If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then the terms $a,b,c$ are in $G.P$ or $A.P$ or $H.P\:\:?$

Then what should be one's approach. Will expanding work here$?$

Any help is greatly appreciated.

Answer 1

As mentioned in comments, you haven't really proved that the equation implies $a, b, c$ are in G.P., you have only shown the other way round.

Here is a hint to address the original problem, $$(ab+bc+ca)^3-abc(a+b+c)^3=(ab-c^2)(bc-a^2)(ca-b^2)$$ Now you can show that if any of the terms on RHS is zero, that implies $a, b, c$ are in G.P.

Answer 2

Macavity has given a good answer for how to go about the original problem. To answer your original question,

Suppose the question was

If $a,b,c$ are non zero real numbers satisfying $$(ab+bc+ca)^3=abc(a+b+c)^3$$ then the terms $a,b,c$ are in $G.P$ or $A.P$ or $H.P\:\:?$

Then what should be one's approach. Will expanding work here$?$

You start by trying to come up with counter-examples for each A.P., G.P. H.P, separately. It's not hard to come up with counter-examples for H.P and A.P. But once you struggle to come up with a counter-example for G.P., you might suspect it is true, so maybe you have a go at proving it is true, i.e. that $a,b,c$ are in G.P. . And then if you struggle too much with this, you go back to trying to find a counter-example. Then go back and forth between looking for a counter-example and trying to prove the affirmative. That's the way I do it anyway...

Answer 3

We can think of the problem in terms of polynomials too. Assume that $a, b, c$ are the roots of the cubic polynomial '$ P(x) = x^3 + ux^2 + vx + w$. We get $$-u = a+b+c$$ $$v = ab+bc+ca$$ $$-w = abc$$ Substituting these in the original condition, we have $$v^3 = (-w)(-u)^3 = wu^3$$ or $$ w = \frac{v^3}{u^3}$$ $$P(x) = x^3 + ux^2 + vx +\frac{v^3}{u^3} = (x^3 + \frac{v^3}{u^3}) + (ux^2 + vx) $$ $$= (x + \frac{v}{u})(x^2 + \frac{v^2}{u^2} - \frac{v}{u}x) + ux(x+\frac{v}{u}) $$ This shows $\frac{-v}{u}$ is a root of the P(x). The product of three roots is $-w = \frac{-v^3}{u^3}$. So the product of the other two roots is $\frac{v^2}{u^2}$. We get that square of one root is product of other two and they must be in a G.P.

Answer 4

Concerning your question, in dealing with relations between three members of a progression for equations such as $ \ (ab+bc+ca)^3 \ = \ abc·(a+b+c)^3 \ \ , \ $ it can be useful to take, say, $ \ b \ $ as the "middle" member and write the other two in terms of it.

For an arithmetic progression with a common difference of $ \ d \ \ , \ $ we have $ \ b - d \ , \ b \ , \ b + d \ \ . \ $ This leads to the putative equation $$ [ \ (b-d)·b \ + \ b·(b+d) \ + \ (b-d)·(b+d) \ ]^3 \ \ =^{?} \ \ (b-d)·b·(b+d)·(3b)^3 $$ $$ \Rightarrow \ \ [ \ 3b^2 \ - \ d^2 \ ]^3 \ \ \neq \ \ 27·b^4·(b^2 - d^2) \ \ . $$ We see at once that this is not an equation, since, as a "binomial-cube", the left side must have four terms when "expanded", including powers of $ \ d \ $ as high as $ \ 6 \ \ , \ $ which are clearly not present on the right side. (The two sides are only equal for $ \ d \ = \ 0 \ \ . \ ] $ The members of a harmonic progression produce a similar situation. With the members $ \ \frac{1}{\beta - d} \ , \ b = \frac{1}{\beta} \ , \ \frac{1}{\beta + d} \ \ , \ $ we would need to have $$ [ \ (\beta-d)^{-1}·\beta^{-1} \ + \ \beta^{-1}·(\beta+d)^{-1} \ + \ (\beta-d)^{-1}·(\beta+d)^{-1} \ ]^3 $$ $$ =^{?} \ \ (\beta -d)^{-1}·\beta^{-1}·(\beta+d)^{-1}·[ \ (\beta -d)^{-1} + \beta^{-1} + (\beta+d)^{-1} \ ]^3 $$ $$ \Rightarrow \ \ \left[ \ \frac{3·\beta}{\beta \ · \ (\beta^2 - d^2)} \ \right]^3 \ \ \neq \ \ \frac{1}{\beta · (\beta^2 - d^2)} \ \ · \ \left[ \ \frac{3·\beta^2 \ - \ d^2}{\beta · (\beta^2 - d^2)} \ \right]^3 \ \ , $$ which is also evidently not an equation for $ \ d \ \neq \ 0 \ \ . $

On the other hand, the members $ \ \frac{b}{r} \ , \ b \ , \ br \ $ of a geometric progression with common ratio $ \ r \ $ produces $$ \left[ \ \frac{b}{r}·b \ + \ b·br \ + \ \frac{b}{r}·br \ \right]^3 \ \ =^{?} \ \ \left( \ \frac{b}{r} \ · \ b \ · \ br \ \right) \ · \ \left[ \ \frac{b}{r} \ + \ b \ + \ br \ \right]^3 $$ $$ \Rightarrow \ \ (b^2)^3 \ · \ \left[ \ \frac{1}{r} \ + \ 1 \ + \ r \ \right]^3 \ \ =^{!} \ \ b^3 \ · \ b^3 · \left[ \ \frac{1}{r} \ + \ 1 \ + \ r \ \right]^3 $$ as you found by doing something also this line (although we can observe the equality or inequalities without expanding the implied multinomials).

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I agree that Macavity's factorization is probably the most satisfying argument for showing that a geometric progression needs to be present among some ordering of the three numbers in order to satisfy the equation (addressing the original problem). I had considered dividing out a factor of $ \ abc \ $ from the equation to obtain $$ \left( \ b \ + \ c \ + \ \frac{b}{a}·c \ \right)·\left( \ a \ + \ c \ + \ \frac{a}{b}·c \ \right)·\left( \ a \ + \ b \ + \ \frac{a}{c}·b \ \right) \ \ = \ \ (a \ + \ b \ + \ c)^3 \ \ . $$ One could make a glib argument that each factor of the left side must match one of the three identical factors on the right side, giving us $ \ \frac{b}{a}·c \ = \ a \ \Rightarrow \ a^2 \ = \ bc \ \ , \ $ and similarly, $ \ b^2 \ = \ ac \ \ , \ c^2 \ = \ ab \ , \ $ so that a geometric progression must be present among some ordering of $ \ a \ , \ b \ , \ c \ \ . $ However, this is only really convincing if $ \ a + b + c \ $ is a prime integer; otherwise, it is difficult to be persuasive that the equality of the two products permits us to "match factors" in that way.