Let $x_1, ..., x_{25}>0$ be such that $\sum_{i=1}^{25}{x_i} = 4350$ and $\sum_{i=1}^{25}{x_i^2} = 757770.25$.
From the first equality alone, we know that at least one of the $x_i$'s must be less than or equal to $\frac{4350}{25}=174$. From the second equality alone, we know that at least one of the $x_i$'s must be less than or equal to $\sqrt{\frac{757770.25}{25}}=174.1$, which is less useful than the first bound. My question is whether we can get a better bound, i.e. to find the least upper bound of $\min{\{x_1, ..., x_{25}\}}$, when we use both equalities together. I appreciate any comments or hints.
Here's some nontrivial improvement at least. Suppose that $m=\min\{x_1,\dots,x_{25}\}\le174$, and write $x_j = m + t_j$. Then $\sum x_j = 4350$ means $\sum t_j = 4350 - 25m$, while $757770.25 = \sum x_j^2 = \sum (m^2 + 2m t_j + t_j^2)$ means \begin{align*} \sum t_j^2 &= 757770.25 - 25m^2 - 2m \sum t_j \\ &= 757770.25 - 25m^2 - 2m(4350 - 25m) = 757770.25 - 8700m + 25m^2. \end{align*} Finally, the fact that the $t_j$ are all nonnegative means that $(\sum t_j)^2 \ge \sum t_j^2$; in other words, we must have $$ (4350 - 25m)^2 - (757770.25 - 8700m + 25m^2) \ge 0. $$ But now the quadratic equation implies that $m \le \dfrac{20880 - 59\sqrt6}{120}$, which is a little less than $172.8$.
Edited to add: as RobPratt pointed out, this bound is actually optimal! If $x_1=\cdots=x_{24}=\dfrac{20880 - 59\sqrt6}{120}$ and $x_{25} = 4350 - 24x_1 = \dfrac{870+59 \sqrt{6}}5$, then $\sum x_j = 4350$ and $\sum x_j^2 = 757770.25$ exactly.