Showing that $a$, $b$, $c$, $d$ are in geometric progression iff $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$ 
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.
 A: 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
$$ (a^2 r + a^2 r^3 + a^2 r^5)^2 = a^4 r^2 (1+r^2 + r^4)^2 $$
A: 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 $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$:

*

*Continued proportion implies $(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

*If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, then prove that $(a^2+b^2+c^2)(d^2+b^2+c^2)=(ab+bc+cd)^2$
$(a^2+b^2+c^2)(d^2+e^2+f^2)=(ad+be+cf)^2$ implies the equality of ratios:

*

*Example 14, Chapter 1, Higher Algebra by Henry Sinclair.

*Prove that $a(x+y+z) = x(a+b+c)$
Also related:

*

*Prove that $(ax + by + cz)^2 \leq (a^2 + b^2 + c^2)(x^2 + y^2 + z^2)$.
A: 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, $\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2}$.
