<s>[INST] 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?   
$$\frac{a_{1}}{a_{2}/a_{1}}+\dfrac{b_{1}}{b_{2}/b_{1}}\neq \frac{a_{1}+b_{1}}{1+\dfrac{(a_{2}+b_{2})-(a_{1}+b_{1})}{a_{1}+b_{1}}}?$$
 [/INST] 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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