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

Question

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)

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

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

Answer 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

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

Answer 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 $(1).$

Answer 3

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 $x=y$).