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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.

2 Answers 2

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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.

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    @Aryabhatta:oops! yes got it now :-)2011-06-08
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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.