Suppose there are several fractions ${\dfrac{a_1}{b_1}}$,…,${\dfrac{a_n}{b_n}}$ and $a_i$ < $b_i$ for 1≤i≤n all a > 0 and b>0
Mediant of the above fractions
M=(a1+...an)/(b1+⋯+bn)
and the weighted mean of the fractions wrt numerator value
A=${w{\scriptscriptstyle 1}\dfrac{a{\scriptscriptstyle 1}}{b{\scriptscriptstyle 1}}}$ + ${w{\scriptscriptstyle 2}\dfrac{a{\scriptscriptstyle 2}}{b{\scriptscriptstyle 2}}}$ + ... n times
$w_i$ = $\dfrac{a_i}{\sum_{i=1}^{n}a_{i}}$
Is it always true that A≥M? Probably yes, whats the proof. please help
Suppose the inequality is true for $n$ fractions. Then we need to prove that $$ \sum_1^{n+1}w_i\frac{a_i}{b_i}\ge\frac{\sum_1^{n+1}a_i}{\sum_1^{n+1}b_i} $$ This is equivalent to $$ (\sum_1^{n}w_j)\sum_1^{n}\frac{w_i}{\sum_1^{n}w_j}\frac{a_i}{b_i}+w_{n+1}\frac{a_{n+1}}{b_{n+1}}\ge\frac{\sum_1^{n}a_i+a_{n+1}}{\sum_1^{n}b_i+b_{n+1}}. $$ Using the inductive hypothesis, $$ (\sum_1^{n}w_j)\sum_1^{n}\frac{w_i}{\sum_1^{n}w_j}\frac{a_i}{b_i}+w_{n+1}\frac{a_{n+1}}{b_{n+1}}\ge (\sum_1^{n}w_j)\frac{\sum_1^n a_i}{\sum_1^n b_i}+w_{n+1}\frac{a_{n+1}}{b_{n+1}} $$ (observe that the weights are the right ones, corresponding to the first $n$ fractions). Using the fact that we know the inequality is true for two fractions, $$ (\sum_1^{n}w_j)\frac{\sum_1^n a_i}{\sum_1^n b_i}+w_{n+1}\frac{a_{n+1}}{b_{n+1}}\ge \frac{\sum_1^{n}a_i+a_{n+1}}{\sum_1^{n}b_i+b_{n+1}}=\frac{\sum_1^{n+1}a_i}{\sum_1^{n+1}b_i}, $$ as desired (in the last formula, the weights are again the right ones).