protestommb

2022-06-25

Ratio Inequality
How can I prove that,
$\frac{{a}_{1}+{a}_{2}+\cdots +{a}_{n}}{{b}_{1}+{b}_{2}+\cdots +{b}_{n}}\le \underset{i}{max}\left\{\frac{{a}_{i}}{{b}_{i}}\right\}$
where $1\le i\le n$, and ${a}_{i}\ne {a}_{j}$ and ${b}_{i}\ne {b}_{j},\mathrm{\forall }i\ne j$
Edit I have figured out that the above assumptions about ${a}_{i}$, and ${b}_{i}$ are not needed.

The ${a}_{i}$ can be arbitrary real numbers, but the ${b}_{i}$ need to be positive. Then

and adding these gives the desired inequality.
If the ${b}_{i}$ are not required to be positive then the inequality must not hold, a counter-example is
$\frac{2-1}{3-2}>max\left\{\frac{2}{3},\frac{-1}{-2}\right\}\phantom{\rule{thinmathspace}{0ex}}.$

Averi Mitchell

Suppose this holds for $n=2$ (prove this base case yourself). Then
$\frac{\left({a}_{1}+...{a}_{k}\right)+{a}_{k+1}}{\left({b}_{1}+...+{b}_{k}\right)+{b}_{k+1}}\le max\left(\frac{{a}_{1}+...+{a}_{k}}{{b}_{1}+...+{b}_{k}},\frac{{a}_{k+1}}{{b}_{k+1}}\right)$

Do you have a similar question?