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Suppose \begin{align*} a+b+c &= 20\\ \frac 1a + \frac 1b + \frac 1c &= 30 \end{align*}

Then find the value of $$ \frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb $$ How i can apply A.M. $\ge$ G.M $\ge$ H.M. in this?

How i can achieve this?

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2  
Shouldn't it be $\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb$? –  Sidharth Iyer Mar 6 '12 at 8:28
    
Why do you think the AM-GM-HM inequality will help? –  Willie Wong Mar 6 '12 at 8:31
    
Even if it does say that, are you sure it's not a typo? Seems odd to have 2 separate $\frac bc$'s –  Mike Mar 6 '12 at 8:35
    
@Mike ya.i think that was a error.There was a typing mistake in book. –  vikiiii Mar 6 '12 at 8:35

3 Answers 3

up vote 7 down vote accepted

I'm guessing that first $\frac bc$ should be $\frac ba$. I don't know what A.M., G.M., or H.M. are. But as for how to come up with a solution, I can see good things will happen if you multiply both equations together.

$(a+b+c)(\frac1a+\frac1b+\frac1c)=a(\frac1a+\frac1b+\frac1c)+b(\frac1a+\frac1b+\frac1c)+c(\frac1a+\frac1b+\frac1c)$

I'm thinking you should be able to take it from there.

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Thanks @Mike . Got it –  vikiiii Mar 6 '12 at 8:34
    
No problem. It looks like someone actually beat me to posting the answer and erased it though. –  Mike Mar 6 '12 at 8:36

$$\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb $$ can be simplified as

$$ \frac{a^2c+b^2c+a^2b+bc^2+ab^2+ac^2}{abc} $$

which is

$$ \frac{\left[ac(a+c)+bc(b+c)+ab(a+b)\right]}{abc}$$

and that is

$$ \frac{\left[ac(20-b)+bc(20-a)+ab(20-c)\right]}{abc}$$

$$ = \frac{\left[20(ab+bc+ca)-3abc\right]}{abc} $$

But

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=30$$ implies

$$ ab+bc+ca=30abc$$

So you can do further simplification to get $597$

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Exploit the innate symmetry! This is solvable purely mechanically, i.e. no lucky guesses are needed. The rational functions are all symmetric functions of $\rm\:a,b,c.\:$ By the Fundamental Theorem of Symmetric Polynomials, every symmetric polynomial is a polynomial in the elementary symmetric polynomials $\rm\:e_i.\:$ There is a simple effective algorithm due to Gauss to compute this representation. Putting all over a common denominator, we are given

$$\rm e_1\ =\ a + b + c$$

$$\rm \frac{e_2}{e_3}\ =\ \frac{ab + bc + ca}{abc}\ =\ c^{-1} + a^{-1} + b^{-1}$$

and we seek

$$\rm \frac{f(a,b,c)}{e_3}\ =\ \frac{a^2 (b+c) + a\: (b^2+c^2) + b^2 c + b c^2 }{abc}$$

Now use Gauss's algorithm to rewrite $\rm\:f\:$ in terms of the elementary symmetric polynomials $\rm\:e_i.\:$ This works as follows: if $\rm\ a^i\ b^j\ c^k\ $ is the highest term w.r.t. lex order $\rm\ a > b > c\ $ then kill the highest term of $\rm\:f\:$ by subtracting $\rm\ e_1^{i-j}\ e_2^{j-k}\ e_3^k\:.\:$ Here, since $\rm\ a^2 b^1 c^0\: $ is the highest term in $\rm\:f,\:$ we subtract from $\rm\:f\:$ the term $\rm e_1^{2-1}\ e_2^{1-0}\ e_3^0\ = (a+b+c) (ab+bc+ca),\:$ leaving $\rm\: -3abc = -3\:e_3.\:$ Thus we infer$\rm\:f = e_1 e_2 - 3\:e_3,\:$ so $\rm\:f/e_3 = e_1 (e_2/e_3) - 3\: =\: 20\cdot 30 - 3\: =\: 597.$

While this problem is so simple that one may be lucky enough to stumble upon this representation, one will rarely be so lucky with problems of even slightly greater complexity.

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