sqing wrote: and k is non- the smallest of them. what do you mean by that

Sorry. Thanks.

Set $x_i=\big[\sqrt{a_i}\big]+1$, WLOG, $x_1$ is the smallest. It suffices to show that $\sum_{1\le i<j \le 25}2x_i x_j+800 \ge 48 \sum_{i=1}^{25} x_i+ 200 x_1^2.$ When $x_i$'s are all equal, the inequality reduced to $x_1^2+3 \ge 3x_1$, which is true for all positive integers. Now fix $x_1,x_2,...,x_{24}$, consider the function $f(x_{25})=\sum_{1\le i<j \le 25}2x_i x_j-800 - 48 \sum_{i=1}^{25} x_i+ 200 x_1^2=ax_{25}+b$ for some positive constant $a$, constant $b$. So when $f(x_{25})$ achieves minimum, $x_{25}=x_1$. Similarly, can fix all $x_i$'s but $x_j(j>1)$, the corresponding function achieves minimum when $x_{j}=x_1$. So the desired inequality holds as we have checked it holds when $x_i$'s are all equal.

Assume WLOG that $a_{1} \geq a_{2} \geq \dots \geq a_{25}=k$. Let $a_{i} = x_{i}^{2} + y_{i}$, where $0 \leq y_{i} \leq 2x_{i}$ and $x_{i}, y_{i} \in \mathbb{Z}$. Then the given inequality is equivalent to proving that \[ x_{1} + x_{2} + \dots + x_{25} \geq \big[\sqrt{x_{1}^{2} + y_{1} + x_{2}^{2} + y_{2} + \dots + x_{25}^{2} + y_{25} + 200(x_{25}^{2} + y_{25})}\big]. \]In order to prove the last inequality it suffices to prove that \[ \sqrt{x_{1}^{2} + y_{1} + x_{2}^{2} + y_{2} + \dots + x_{25}^{2} + y_{25} + 200(x_{25}^{2} + y_{25})} < x_{1} + x_{2} + \dots + x_{25} + 1. \]After taking square of both sides and using the inequality $2(x_{1} + x_{2} + \dots + x_{25}) + 1 > y_{1} + y_{2} + \dots + y_{25}$ we note that proof can be completed if one proves that \[ 100(x_{25}^{2} + y_{25}) < \sum_{1 \leq i < j \leq 25}x_{i}x_{j}. \] If $x_{25}=0$ then the last inequality obviously holds for all except for the case $x_{2}=x_{3} = \dots = x_{25}=0$. However, for this case $a_{2}=a_{3}=\dots=a_{25}=0$ and direct verification of the original inequality shows that (in)equality holds. We may now assume that $x_{25} \geq 1$. If $x_{1} = x_{2} = \dots = x_{25} = 1$ then the goal is to prove that $25 \geq \big[\sqrt{a_{1} + a_{2} + \dots + 201 a_{25}} \big]$ which is true as $a_{i} \leq 3$ implying that $a_{1} + a_{2} + \dots + 201 a_{3} \leq 675 < 676 = 26^{2}$. Otherwise, $x_{1} \geq 2$. We note that there are $300$ summands on RHS and obviously $x_{i}x_{j} \geq x_{25}^{2}$. Furthermore, $x_{i}x_{j} + x_{t}x_{s} \geq (x_{i}+x_{t})x_{25} \geq 2x_{25} \geq y_{25}$ (i.e. each two summands on RHS are greater than or equal to $y_{25}$). As $x_{1}x_{2} + x_{1} x_{3} \geq 2x_{1}x_{25} \geq 4x_{25} \geq 2+ 2x_{25} > y_{25}$ we can combine summands on RHS to easily prove that $100(x_{25}^{2} + y_{25}) < \sum_{1 \leq i < j \leq 25}x_{i}x_{j}$ holds.

By smoothing it suffices to prove the inequality in the case where $a_i=(x_i+1)^2-1=x_i^2+2x_i$, where $x_i$ is a nonnegative integer. The inequality then becomes $$x_1+\cdots+x_{25} \geq \lfloor \sqrt{x_1^2+\cdots+x_{25}^2+2x_1^2+\cdots+2x_{25}^2+200k}\rfloor \iff (x_1+\cdots+x_{25}+1)^2>x_1^2+\cdots+x_{25}^2+2x_1^2+\cdots+2x_{25}^2+200k.$$By expansion, this is equivalent to proving that $$\sum_{i<j} x_ix_j \geq 100k.$$WLOG let $k=x_1^2+2x_1$. Then the above inequality follows from $x_ix_j \geq x_1^2 \geq \tfrac{1}{3}(x_1^2+2x_1)$. $\blacksquare$