Proof of cauchy's third inequality

Question

Show that for positive $h_i, b_i$ where $h_i / b_i \leq h_j/b_j$ if $i < j$ that the following inequality holds

$$\frac{h_1}{b_1} \leq \frac{\sum_i^n h_i}{\sum_i^nb_i} \leq \frac{h_n}{b_n}$$

My proof uses the mean value theorem to prove this result, but I am looking for alternative proofs. Specifically I wonder if it's possible to prove this with some variation of the rearrangement inequality.

This is the “mediant inequality” – duplicate target found with Approach0 — Martin R, Sep 26 '23 at 08:08

score 2 · Accepted Answer · answered Sep 26 '23 at 00:17

\begin{align} \frac{h_i}{b_i}\le\frac{h_n}{b_n}\ \ \text{ for all }\ i&\Rightarrow\frac{h_i}{h_n}\le\frac{b_i}{b_n}\ \ \text{ for all }\ i\\ &\Rightarrow\sum_{i=1}^n\frac{h_i}{h_n}\le\sum_{i=1}^n\frac{b_i}{b_n}\\ &\Rightarrow\frac{\sum_\limits{i=1}^nh_i}{\sum_\limits{i=1}^nb_i}\le\frac{h_n}{b_n} \end{align} Likewise, \begin{align} \frac{h_1}{b_1}\le\frac{h_i}{b_i}\ \ \text{ for all }\ i&\Rightarrow\frac{b_i}{b_1}\le\frac{h_i}{h_1}\ \ \text{ for all }\ i\\ &\Rightarrow\sum_{i=1}^n\frac{b_i}{b_1}\le\sum_{i=1}^n\frac{h_i}{h_1}\\ &\Rightarrow\frac{h_1}{b_1}\le\frac{\sum_\limits{i=1}^nh_i}{\sum_\limits{i=1}^nb_i} \end{align}

That's great, no need for anything fancy at all. Thanks! – dmh Sep 26 '23 at 00:18 — dmh, Sep 26 '23 at 00:18

Proof of cauchy's third inequality

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