It's clear that if $A$ is a finite set of non-negative reals, then $1+\sum_{x\in A} x\le \prod_{x\in A}(1+x)$. $\sum_{i=1}^n a_i = n+ \sum_{i=1}^{n}(a_i -1) = n+ \sum_{a_i\le1}(a_i -1) +\sum_{a_i>1}(a_i -1)$ $\le n + \sum_{a_i>1}(a_i -1) \le (n-1)+ \prod_{a_i>1}a_i = (n-1) + \prod_{i=1}^{n}{\max{\{1,a_i\}}}$

Let $b_{k}=\max \{ 1,a_{k}\}$, so that \[\sum_{k=1}^{95}a_{k}\leq \sum_{k=1}^{95}b_{k},\ 94+\prod_{k=1}^{95}\max \{ 1,a_{k}\}=94+\prod_{k=1}^{95}b_{k}\] So we will be done if we can prove that \[\sum_{k=1}^{95}b_{k}\leq 94+\prod_{k=1}^{95}b_{k}\] knowing that $b_{k}\geq 1$ for each $k$. Of course, this suggests \[\sum_{k=1}^{n}b_{k}\leq n-1+\prod_{k=1}^{n}b_{k}\] is true for any positive integer $n$. We show that this is in fact true, using induction. The base case is clearly true. For the inductive step, \[n+b_{1}b_{2}b_{3}\cdots b_{n+1}=n+b_{1}b_{2}b_{3}\cdots b_{n}+b_{1}b_{2}b_{3}\cdots b_{n}(b_{n+1}-1)\] \[\geq 1+(b_{1}+b_{2}+b_{3}+\cdots+b_{n})+b_{1}b_{2}b_{3}\cdots b_{n}(b_{n+1}-1)\geq b_{1}+b_{2}+b_{3}+\cdots+b_{n+1}\] where we have simply used $b_{n+1}\geq 1,\ b_{1}b_{2}\cdots b_{n}\geq 1$ for the last step. This proves the inequality for any positive integer $n$; in particular, it works for $n=95$.