<s>[INST] 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.
 [/INST] 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
$$ (a^2 r + a^2 r^3 + a^2 r^5)^2 = a^4 r^2 (1+r^2 + r^4)^2 $$
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