# Ratio Inequality How can I prove that, a <mrow class="MJX-Te

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.
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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
$\frac{2-1}{3-2}>max\left\{\frac{2}{3},\frac{-1}{-2}\right\}\phantom{\rule{thinmathspace}{0ex}}.$
###### Did you like this example?
Averi Mitchell
Suppose this holds for $n=2$ (prove this base case yourself). Then
$\frac{\left({a}_{1}+...{a}_{k}\right)+{a}_{k+1}}{\left({b}_{1}+...+{b}_{k}\right)+{b}_{k+1}}\le max\left(\frac{{a}_{1}+...+{a}_{k}}{{b}_{1}+...+{b}_{k}},\frac{{a}_{k+1}}{{b}_{k+1}}\right)$