Showing that $a$, $b$, $c$, $d$ are in geometric progression iff $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

Question

If the real numbers $a$, $b$, $c$, $d$ are in geometric progression, show that $$ \left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} $$ Prove that the converse also holds.

The simplest way I could think of is making/assuming a general G.P

Let $r$ be the common ratio and $a$ be the first term, then $$b=a r, c=a r^{2}, d=a r^{3}$$ after multiplying a lot of terms, many times, (skipping significant steps here so it is more readable)

$\begin{aligned} \mathrm{LHS} &=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}\right)\left(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right) \\ &=a^{2}\left(1+r^{2}+r^{4}\right) \cdot a^{2} r^{2}\left(1+r^{2}+r^{4}\right) \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} \\ &=\left(a^{2} r+a^{2} r^{3}+a^{2} r^{5}\right)^{2} \\ &=\left(a \cdot a r+a r \cdot a r^{2}+a r^{2} \cdot a r^{3}\right)^{2} \\ &=(a b+b c+c d)^{2}=R H S \end{aligned}$

Once after expanding each term, and getting no where, I realised that I had already gotten the answer, all I had to do was take the terms inside the and then I had the RHS. But it all was so tedious and took multiple attempts. Though, this method guarantees that converse holds,

Can it be done more elegantly?

Edit: I am in highschool (and just a little more interested in math) So I don't have knowledge about the much spoken Cauchy's identity.

You might check some older related posts on this site, such as: Continued proportion implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$ and Example 14, Chapter 1, Higher Algebra by Henry Sinclair.. I found these posts by searching in SearchOnMath and ... — Martin Sleziak, Jul 08 '20 at 13:40
.. in Approach0. For general advice on searching, see: How to search on this site? — Martin Sleziak, Jul 08 '20 at 13:40
@Martin Sleziak, I tried to, but I think the search does not search something in $ sign.. I couldn't find that. Thanks! — Ashley, Jul 08 '20 at 13:42
One thing to ask is also whether you're familiar with Cauchy–Schwarz inequality. In particular, whether you know whether in which cases we actually have equality. — Martin Sleziak, Jul 08 '20 at 13:47
@Martin Sleziak, When I delete the question it shows there is one answer and when I undelete it, is shows 0. WHY ?????? — Ashley, Jul 08 '20 at 13:48
No I am, not but I found about Lagrange's identity in a comment in the second link, and that's helpful I'm trying to understand that currently. — Ashley, Jul 08 '20 at 13:49
The converse does not always hold: if $a=b=c=0$ and $d\ne 0$, then we have equality but the numbers are not in geometric progression. — rogerl, Jul 08 '20 at 14:54
This question is from RS Aggarwal Mathematics Textbook Class 11. See also the next question https://math.stackexchange.com/questions/4214137/if-a-b-c-d-are-in-g-p-prove-that-they-are-in-the-same-place-as-if-they-we — , Aug 01 '21 at 08:20

score 4 · Answer 1 · answered Jul 08 '20 at 13:58

4

Left side is $$(a^2 + (ar)^2 + (ar^2)^2)((ar)^2 + (ar^2)^2 + (ar^3)^2) = a^4 r^2 (1 + r^2 + r^4)^2$$ Right side is $$ (a^2 r + a^2 r^3 + a^2 r^5)^2 = a^4 r^2 (1+r^2 + r^4)^2 $$

answered Jul 08 '20 at 13:58

Robert Israel

448,999

How did you see that in two steps? – Ashley Jul 08 '20 at 13:59
3

If I may add a minor comment, this is only one implication. The question is formulated as "iff", so it still remains to show that if the equality holds, then $a$, $b$, $c$, $d$ are in a geometric progression. – Martin Sleziak Jul 08 '20 at 14:34
2

@Ashley First extract the power of $a$, then see what powers of $r$ appear. – Robert Israel Jul 08 '20 at 14:53
-1 where is the other way – nonuser Jul 09 '20 at 07:32
As others have mentioned it's from Cauchy-Schwarz. – Robert Israel Jul 09 '20 at 12:30

Martin Sleziak · Accepted Answer · 2020-07-09T08:06:00.480

^{I am pretty sure that I am basically repeating stuff which was already said in other posts on this site. However, the other questions that I was able to find were about one implication, this one is about equivalence. Moreover, the OP explicitly mentioned that they prefer answers at highschool level. Both these reasons make this question a bit different. I have also included some collection of other posts with similar topic at the end of this answer. The answer is CW, I made it community wiki mainly to encourage other users to add more posts to that list, if they found such posts.$\newcommand{\abs}[1]{\lvert#1\rvert}$}

Let us consider the vectors $\vec x=(a,b,c)$ and $\vec y=(b,c,d)$. Their dot product is $\vec x\cdot\vec y = ab+bc+cd$. So the given equality can be rewritten as \begin{align*} (a^2+b^2+c^2)(b^2+c^2+d^2) &= (ab+bc+cd)^2 \\ \abs{\vec x}^2\cdot\abs{\vec y}^2 &= \abs{\vec x\cdot\vec y}^2\\ \abs{\vec x}\cdot\abs{\vec y} &= \abs{\vec x\cdot\vec y} \\ \abs{\vec x}\cdot\abs{\vec y} &= \abs{\vec x}\cdot\abs{\vec y}\cdot\cos\varphi \end{align*} where the $\varphi$ is the angle between $\vec x$ and $\vec y$.

If we assume that the vectors $\vec x$ and $\vec y$ are non-zero, then the happens if and only if $\cos\varphi=1$. This means that one of the vectors is multiple of the other vector.

If we have that $\vec y=k\vec x$, then we get $(b,c,d)=k(a,b,c)$, i.e., $b=ka$, $c=kb$ and $d=kc$. So this gives us that the given for numbers are in a geometric progression and we are done.

It was already pointed out in a comment that we have to disallow zeroes in order to get equivalence. (So we're trying to prove the equivalence for non-zero numbers $a$, $b$, $c$, $d$. For such numbers, we get that $\vec x \ne \vec0$, $\vec y \ne \vec0$; we have used this assumption above.)

Remark 1. Notice that basically the same argument can be used to show that for non-zero numbers $a$, $b$, $c$, $d$, $e$, $f$ we have $$(a^2+b^2+c^2)(d^2+e^2+f^2)=(ad+be+cf)^2$$ if and only if $\frac da=\frac eb=\frac fc.$

Remark 2. What I have done here is basically using Cauchy–Schwarz inequality. (More precisely, looking at the case when there is equality in Cauchy–Schwarz inequality.) As the OP mentioned in the comments that they do not know this inequality, I opted for the description using vectors.

Here are some other posts about this problem (or at least about one implication)

Geometric progression implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$:

$(a^2+b^2+c^2)(d^2+e^2+f^2)=(ad+be+cf)^2$ implies the equality of ratios:

Also related:

Prove that $(ax + by + cz)^2 \leq (a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$.

score 0 · Answer 3 · answered Aug 01 '21 at 07:57

Q. If $a, b, c, d$ are in G.P., prove that $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} .$

Sol. Let $r$ be the common ratio of the G.P. $a, b, c, d$.

Then $b=a r ; c=a r^{2}$ and $d=a r^{3}$.

$\begin{aligned} \therefore \quad \text { L.H.S. } &=\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right) \\ &=\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}\right)\left(a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\right) \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} \\ \text { And, } R \cdot H . S_{.} &=(a b+b c+c d)^{2} \\ &=\left(a^{2} r+a^{2} r^{3}+a^{2} r^{5}\right)^{2} \\ &=a^{4} r^{2}\left(1+r^{2}+r^{4}\right)^{2} . \end{aligned}$

Hence, $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$.

Showing that $a$, $b$, $c$, $d$ are in geometric progression iff $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

3 Answers3