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