# Why does the sum of a division applied to individual items not equal the division applied to the sum of those items?

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?

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

-
@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

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

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

-
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 $(1).$

-
i've updated the question to make it clearer – Stafford Williams Mar 3 '11 at 14:05

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 $x=y$).

-
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