Given:
$\frac{a_{1}}{b_{1}}$ $\le$ $\frac{a_{2}}{b_{2}}$ $\le$ ... $\le$ $\frac{a_{n}}{b_{n}}$ with all ${b_{i}}$ being positive
Proof by using induction that:
$\frac{a_{1}}{b_{1}}$ $\le$ $\frac{a_{1} + a_{2} + ... + a_{n}}{b_{1} + b_{2} + ... + b_{n}}$ $\le$ $\frac{a_{n}}{b_{n}}$
Attempt:
Let the base case be n = 2: $\frac{a_{1}}{b_{1}}$ $\le$ $\frac{a_{1} + a_{2}}{b_{1} + b_{2}}$ $\le$ $\frac{a_{2}}{b_{2}}$ (This can be quite easily proven by proving the inequalities seperately and getting rid of the denominator)
Assume that it holds true for n = k, then it should hold true for n = k + 1
How would I go about proving this? I know there is already a topic out there (shorter proof of generalized mediant inequality?), but induction isn't used there. This problem appears in Calculus Deconstructed, page 7.
I think this proof of induction should have a previous lemma similar to the following:
Lemma Let $a/b$ and $c/d$ reduced fractions where all $a,b,c,d$ are rational numbers and such that $a/b < c/d$. Let $e/f$ be a reduced fraction such that $a/b < e/f$ and $c/d < e/f$. Then, \begin{equation} \frac{a}{b} \oplus \frac{e}{f} \le \frac{c}{d} \oplus \frac{e}{f}, \end{equation} where $u/v \oplus w/x = (u + w)/(v + x)$.
If this was true, then I think we can easily solve the induction step. Now follows a sketch of the proof.
Assume that the inequality \begin{equation} \frac{a_1}{b_1} \le \frac{a_1 + \dots + a_k}{b_1 + \dots + b_k} \le \frac{a_k}{b_k} \end{equation} holds for all $k\in[1,n]$. Now, does it still hold for $k+1$? Now we apply the lemma above to the inequality by using $a_{k+1}/b_{k+1}$ which, as per your formulation, we have that
\begin{equation} \frac{a_1}{b_1} \le \frac{a_{k+1}}{b_{k+1}}, \qquad \frac{a_k}{b_k} \le \frac{a_{k+1}}{b_{k+1}} \end{equation}
Then, if we use the summation of the mediant of the fractions
\begin{equation} \frac{a_1}{b_1} \oplus \frac{a_{k+1}}{b_{k+1}} \le \frac{a_1 + \dots + a_k}{b_1 + \dots + b_k} \oplus \frac{a_{k+1}}{b_{k+1}} \le \frac{a_k}{b_k} \oplus \frac{a_{k+1}}{b_{k+1}}. \end{equation}
Finally, operate the fractions with $\oplus$,
\begin{equation} \frac{a_1 + a_{k+1}}{b_1 + b_{k+1}} \le \frac{a_1 + \dots + a_k + a_{k+1}}{b_1 + \dots + b_k + b_{k+1}} \le \frac{a_k + a_{k+1}}{b_k + b_{k+1}} \end{equation}
and apply the base case ($n=2$) to the left and right terms of the inequality using the assumption that $a_1/b_1 < a_2/b_2 < \dots < a_k/b_k < a_{k+1}/b_{k+1}$:
\begin{equation} \frac{a_1}{b_1} \le \frac{a_1 + a_{k+1}}{b_1 + b_{k+1}} \le \frac{a_1 + \dots + a_k + a_{k+1}}{b_1 + \dots + b_k + b_{k+1}} \le \frac{a_k + a_{k+1}}{b_k + b_{k+1}} \le \frac{a_{k+1}}{b_{k+1}}. \end{equation}
This concludes the proof.