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I hope the simplicity of this problem doesn't offend anyone's intelligence, however i'm unsure where else I should post this, and fairly certain its due to some kind of math rule (or just my stupid mistake);

I'm trying to figure out why the figures in the far right boxes** below are not equal. How is the sum of the results of the division of individual items larger than when the division is applied to the already summed items?

** Here's a link to the model i've constructed (which did nothing but confuse, see below)

enter image description here

EDIT Unfortunately it looks like i've mislead with the 0.05 looking like an average (hence the /2 in comments & answers), when in fact it is not. I've got my question into an equation now, so hopefully this is clearer (and yeah, my bad that this equation probably looks nothing like the original model, apologies)

Revised Question:

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

Then

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

So why when $a_2/a_1 \neq b_2/b_1 , a_1 \neq b_1$

Then

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

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@Stafford - it is a bit unclear why you think they would be equal? It looks like you think $a_1/(1+a_2)+b_1/(1+b_2) = (a_1+a_2)/(1+a_2 b_2/2)$? (where $a_1,a_2,b_1,b_2$ are in this case 1,0,1,0.1) –  Juan S Mar 3 '11 at 10:31
    
There's no rule in effect here, there's an absence of a rule: there's no reason the thing you think should be true should actually be true, and this is a counterexample. –  Qiaochu Yuan Mar 3 '11 at 12:25
    
@Qwirk i didn't intend the 2 to be there, i've expanded the question, hopefully it helps clarify –  Stafford Williams Mar 3 '11 at 14:00
    
@Stafford Williams: condition $a_1\neq b_1$ is not required, does it? –  Américo Tavares Mar 3 '11 at 17:10
    
@Américo hmm, should have been $a_2 \neq b_2$. When the tax component is equal, so are the formulas. –  Stafford Williams Mar 8 '11 at 2:27

3 Answers 3

up vote 1 down vote accepted

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

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

enter image description here

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érico how does the $1+$ get eliminated and replaced with $a_1+b_1$ on the right-hand side of the second line of your proof? –  Stafford Williams Mar 5 '11 at 5:41
    
@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 $$\frac{(a_{1}+b_{1})\cdot (a_{1}+b_{1})}{ (a_{1}+b_{1})\cdot 1+(a_{1}+b_{1})\cdot\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}=\frac{% \left( a_{1}+b_{1}\right) ^{2}}{a_{1}+b_{1}+(a_{2}+b_{2})-(a_{1}+b_{1})}$$ –  Américo Tavares Mar 5 '11 at 11:25

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 clearer –  Stafford Williams Mar 3 '11 at 14:05

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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in the revised question numbers $a_1,a_2,b_1,b_2$ may be negative. –  Américo Tavares Mar 3 '11 at 17:01
    
@Americo: Actually the first proof only assumes $a_1 b_1 \neq 0$. Edited. btw +1 to you answer :-) –  Aryabhata Mar 3 '11 at 17:26
    
+1. You also assume $xy\neq 0$ ($a_2,b_2\neq 0$) in the denominators of the 1st fraction. –  Américo Tavares Mar 3 '11 at 17:56
    
@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. –  Aryabhata Mar 3 '11 at 17:57

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