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?   
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$$
 A: 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 $(1).$
A: 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 $x=y$).
A: 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
$$\frac{1}{1/1}+\frac{1}{1.1/1}\neq \frac{2}{1+\frac{(1+1.1)-\left( 1+1\right) }{2%
}}=\frac{2}{1.05}\Leftrightarrow \frac{1}{1}\neq \frac{1.1}{1}.$$

