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When $a_2/a_1 = b_2/b_1$, $a_1 \neq b_1$, we have

$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}= \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}.$

So why when $a_2/a_1 \neq b_2/b_1 , a_1 \neq b_1$ we don't have a similar equality?

$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$

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    @Stafford, 3. In general you would have to solve for $x$ the equation $\frac{a_{1}}{1+a_{2}}+\frac{b_{1}}{1+b_{2}}=\frac{a_{1}+b_{1}}{1+x}$ ($a_{2}$, $b_{2}$, taxes, $a_{1}$, $b_{1}$, revenues and $x$, the average tax). If $a_{2}=b_{2}$, then $x=a_{2}=b_{2}$ and now for this new equation indeed "When the tax component is equal, so are the formulas".2011-03-08

3 Answers 3

1

Because for $a_{1},a_{2},b_{1},b_{2}\neq 0$ we have the following equivalent inequalities:

$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$

This can be shown as follows:

$\frac{a_{1}}{\dfrac{a_{2}}{a_{1}}}+\dfrac{b_{1}}{\dfrac{b_{2}}{b_{1}}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}$

$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{% \left( a_{1}+b_{1}\right) ^{2}}{a_{1}+b_{1}+(a_{2}+b_{2})-(a_{1}+b_{1})}$

$\Leftrightarrow \frac{a_{1}^{2}}{a_{2}}+\frac{b_{1}^{2}}{b_{2}}\neq \frac{% \left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$

$\Leftrightarrow \frac{a_{1}^{2}b_{2}+a_{2}b_{1}^{2}}{a_{2}b_{2}}\neq \frac{% \left( a_{1}+b_{1}\right) ^{2}}{a_{2}+b_{2}}$

$\Leftrightarrow \left( a_{1}^{2}b_{2}+a_{2}b_{1}^{2}\right) \left( a_{2}+b_{2}\right) \neq \left( a_{1}+b_{1}\right) ^{2}a_{2}b_{2}$

$\Leftrightarrow a_{1}^{2}b_{2}^{2}+a_{2}^{2}b_{1}^{2}-2a_{2}b_{2}a_{1}b_{1}\neq 0$

$\Leftrightarrow \left( a_{1}b_{2}-a_{2}b_{1}\right) ^{2}\neq 0$

$\Leftrightarrow a_{1}b_{2}\neq a_{2}b_{1}$

$\Leftrightarrow \frac{a_{2}}{a_{1}}\neq \frac{b_{2}}{b_{1}}.$


Your numerical case seems to be for $a_{1}=b_{1}=a_{2}=1,b_{2}=1.1$ for which we have

$\frac{1}{1/1}+\frac{1}{1.1/1}\neq \frac{2}{1+\frac{(1+1.1)-\left( 1+1\right) }{2% }}=\frac{2}{1.05}\Leftrightarrow \frac{1}{1}\neq \frac{1.1}{1}.$

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    @Stafford: If you multiply the numerator and denominator of $\frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}$ by $a_1+b_1$ you get $\frac{(a_{1}+b_{1})\cdot (a_{1}+b_{1})}{ (a_{1}+b_{1})\cdot 1+(a_{1}+b_{1})\cdot\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}=\frac{% \left( a_{1}+b_{1}\right) ^{2}}{a_{1}+b_{1}+(a_{2}+b_{2})-(a_{1}+b_{1})}$2011-03-05
2

I'm not sure where the figures are coming from, but an interpretation is the inequality (a,b,c > 0) $\frac{a}{a+b} + \frac{a}{a+c} \ge \frac{2a}{a+ (b+c)/2}, \quad (1)$

which follows from

$((a+b)-(a+c))^2 \ge 0$

and so

$(a+b)^2+(a+c)^2 \ge 2(a+b)(b+c).$

Therefore, on adding $2(a+b)(b+c)$ to both sides,

$((a+b)+(a+c))^2 \ge 4(a+b)(b+c)$

and dividing both sides by $((a+b)+(a+c))(a+b)(b+c)$ and multiplying by $a$ will give us $(1).$

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    i've updated the question to make it clearer2011-03-03
1

Your observation is correct!

Let $\displaystyle \frac{a_2}{a_1} = x$ and $\displaystyle \frac{b_2}{b_1} = y$.

Then suppose the expressions you have are equal.

We get

$\frac{a_1}{x} + \frac{b_1}{y} = \frac{(a_1 + b_1)^2}{xa_1 + yb_1}$

This gives us

$(ya_1 + xb_1)(xa_1 + yb_1) = xy(a_1 + b_1)^2$

Which gives us, after some algebra and cancelling $\displaystyle a_1 b_1$,

$x^2 + y^2 = 2xy$

i.e.

$ (x-y)^2 = 0 $

and thus,

$x = y$

So if the two expressions you have are equal, then it is necessarily true that $\displaystyle \frac{a_2}{a_1} = \frac{b_2}{b_1}$.

In fact you will always have (for positive reals)

$\frac{a_1}{x} + \frac{b_1}{y} \ge \frac{(a_1 + b_1)^2}{xa_1 + yb_1}$

the equality occurring iff $\displaystyle x = y$

Another (possibly quicker than the above) way to see this inequality is to apply Cauchy Schwarz to $\displaystyle (\sqrt{\frac{a}{x}}, \sqrt{\frac{b}{y}})$ and $\displaystyle (\sqrt{ax}, \sqrt{by})$, equality occuring only when these are linearly dependent (implying $x=y$).

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    @Americo: The OP has also specified $x$ and $y$ to be in the denominators, just like $a_1$ and $b_1$, so I am not actually making that assumption, OP is.2011-03-03