Given $n$ positive real numbers $a_1\geq a_2,\cdots\geq a_n>0$. Assume $\sum_{i=1}^n a_i=b_1, \sum_{i=1}^n a_i^2=b_2$. I want to find an upper bound on $\frac{a_1}{a_n}$, and the condition when this upper bound is achieved.
My intuition is that the upper bound is achieved either $a_1=a_2=,\cdots=a_{n-1}>a_n$ or $a_1>a_2=a_3=\cdots a_n$, but I don't know how to prove it.
Some ideas: if for any fixed small $a_n$, the equations $a_1+\cdots+a_{n-1}=b_1-a_n, a_1^2+\cdots+a_{n-1}^2=b_2-a_n^2$ has solutions, then one can let $a_n\to 0$, and the $\frac{a_1}{a_n}\to\infty$. To prevent this from happening, we need some conditions on $b_1, b_2$ so that $a_n$ has a positive lower bound. Specifically, one can use the following inequality
\begin{align}\frac{a_1+\cdots+a_{n-1}}{n-1} \leq \sqrt{\frac{a_1^2+\cdots+a_{n-1}^2}{n-1}}\end{align}
We plug in $a_1+\cdots+a_{n-1}=b_1-a_n$ and $a_1^2+\cdots+a_{n-1}^2=b_2-a_n^2$, after some calculations, we get $na_n^2-2b_1a_n+b_1^2-(n-1)b_2\leq 0$. Therefore, as long as $b_1^2-(n-1)b_2\leq 0$, then $a_n$ will have a positive lower bound. Then, my question is when $b_1^2-(n-1)b_2\leq 0$, what is the maximum value of $\frac{a_1}{a_n}$ and when the maximum is achieved?