<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] 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 $(1).$
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