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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.

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2 Answers 2

up vote 5 down vote accepted

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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If we put $\frac{b}{a}=\frac{d}{c}=k$ then $b=ak$ and $d=ck$ but $c=kb$ --- how ? –  Quixotic Jun 8 '11 at 18:35
    
If $a,b,c,d$ are in continued proportion then there exists $k$ such that $\frac{b}{a}=\frac{c}{b}=\frac{d}{c}=k$. –  Chris Taylor Jun 8 '11 at 18:36
    
I don't (properly) understand how could we proof this? –  Quixotic Jun 8 '11 at 18:38
    
@Deb: That is the definition, no proof is required. –  Aryabhata Jun 8 '11 at 18:42
    
@Aryabhatta:oops! yes got it now :-) –  Quixotic Jun 8 '11 at 18:50
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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.

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