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

Question

I am trying to find a tricky way to proof these:

If $a,b,c,d$ are in continued proportion, prove that $$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2 ,$$ This result could be extended to $$(a^2+b^2+c^2+d^2)(b^2+c^2+d^2+e^2)=(ab+bc+cd+de)^2$$ when $a,b,c,d,e$ are in continued proportion.

The standard way for solving them could be putting $\frac{a}{b}=\frac{c}{d}=k$ then followed by substitution and which is followed by tedious algebraic manipulations,but that is not what I am looking for could these be solved in a less easy way using some other algebraic method/tricks? Please explain.

Answer 1

You know that $b=ka$, $c=kb$ etc so the lhs can be rewritten

$$(a^2+b^2+c^2)(k^2a^2+k^2b^2+k^2c^2) = k^2(a^2+b^2+c^2)^2$$

and the rhs can be written

$$(ka^2 + kb^2 + kc^2)^2 = k^2(a^2+b^2+c^2)^2$$

and you're done. The same trick works for the second example.

Answer 2

In fact the converse is also true.

This is the equality case of the Cauchy Schwarz inequality under the Euclidean Norm.

Take $\mathrm{x} = (a,b,c)$ and $\mathrm{y} = (b,c,d)$.

Geometrically, the ratio ($\sqrt{\frac{RHS}{LHS}}$) gives the cosine of the angle between $\mathrm{x}$ and $\mathrm{y}$ and is $1$ only when they are collinear (or linearly dependent).

It applies to higher dimensions too.