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If H is the harmonic mean between $a$ and $b$,then show that $$\frac{1}{H-a}+\frac{1}{H-b} = \frac{1}{a} + \frac{1}{b}$$ and $$\frac{H+a}{H-a}+\frac{H+b}{H-b} = 2$$

I substituted $\displaystyle H = \frac{2ab}{a+b}$, then tried some algebraic manipulation, but I am not getting there.

Are they even valid? If yes, could somebody give me some ideas how to approach these?

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Your method should work. If you show your work here, we can point your error. Other possibilites are: Notice that the right-hand-side of the first equation is $2/H$. Multiply by all three denominators, subtract $2H^2$ and solve for $H$. –  Phira May 13 '11 at 17:07
    
If you are worried about validity, you can just try some values. I put it into Excel and they appear to be. It's not a proof, but it can be a disproof. –  Ross Millikan May 13 '11 at 17:14
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They are not valid if $a=b$. –  Aryabhata May 13 '11 at 17:23
    
Note what @Aryabhatta says. Whatever he says will always be useful. –  user9413 May 13 '11 at 17:37
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1 Answer

up vote 5 down vote accepted

I don't know why you aren't getting this. You have \begin{align*} \frac{1}{H-a} + \frac{1}{H-b} &= \frac{1}{\frac{2ab}{a+b} -a} + \frac{1}{\frac{2ab}{a+b}-b} \\ &= \frac{a+b}{ab-a^{2}} + \frac{a+b}{ab-b^{2}} \\ &= (a+b) \cdot \biggl[ \frac{1}{a(b-a)} + \frac{1}{b(a-b)}\biggr] \\ \end{align*}

I think this should help you out. Try taking the factor $(a-b)$ common. For the next problem also again substitute the value of $H$ and try doing manipulations. Be patient, and be careful, with your calculations, you shall arrive at the result.

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I have tried this far before,but I don't understand what exactly meant by taking the factor $(a-b)$ common? –  Max May 13 '11 at 18:56
    
1:Just i used a pencil and paper ... I do get it from there thanks. –  Max May 13 '11 at 19:03
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