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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.

Tomyy
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    Call $A_k= a_1+a_2+\cdots +a_k$ and $B_k= b_1+b_2+\cdots +b_k$. Your induction hypothesis is then $\frac{a_1}{b_1} \le \frac{A_n}{B_n} \le \frac{a_n}{b_n}$ which you can extend from $n=k$ to $n=k+1$ using your "$n=2$ easily proven" result. (I would have thought the base case was $n=1$) – Henry Feb 21 '23 at 11:28

1 Answers1

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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.

Lluís Alemany-Puig
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