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.

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.

kuncwadi17

Beginner2022-06-26Added 16 answers

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

$\begin{array}{}\text{(*)}& {a}_{j}\le {b}_{j}\cdot \left\{\underset{i}{max}\frac{{a}_{i}}{{b}_{i}}\right\}\phantom{\rule{1em}{0ex}}\text{for}j=1,\dots ,n\end{array}$

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

$\frac{2-1}{3-2}>max\{\frac{2}{3},\frac{-1}{-2}\}\phantom{\rule{thinmathspace}{0ex}}.$

$\begin{array}{}\text{(*)}& {a}_{j}\le {b}_{j}\cdot \left\{\underset{i}{max}\frac{{a}_{i}}{{b}_{i}}\right\}\phantom{\rule{1em}{0ex}}\text{for}j=1,\dots ,n\end{array}$

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

$\frac{2-1}{3-2}>max\{\frac{2}{3},\frac{-1}{-2}\}\phantom{\rule{thinmathspace}{0ex}}.$

Averi Mitchell

Beginner2022-06-27Added 8 answers

Suppose this holds for $n=2$ (prove this base case yourself). Then

$\frac{({a}_{1}+...{a}_{k})+{a}_{k+1}}{({b}_{1}+...+{b}_{k})+{b}_{k+1}}\le max(\frac{{a}_{1}+...+{a}_{k}}{{b}_{1}+...+{b}_{k}},\frac{{a}_{k+1}}{{b}_{k+1}})$

$\frac{({a}_{1}+...{a}_{k})+{a}_{k+1}}{({b}_{1}+...+{b}_{k})+{b}_{k+1}}\le max(\frac{{a}_{1}+...+{a}_{k}}{{b}_{1}+...+{b}_{k}},\frac{{a}_{k+1}}{{b}_{k+1}})$

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