Trying to prove a simple (?) inequality [Sorry, but I couldn't come up with a better title...]
I'm trying to prove that $$\prod_{i=1}^n (1+a_i) \leq 1 + 2\sum_{i=1}^n a_i$$ if all $a_i$ are non-negative reals with $\sum_{i=1}^n a_i \leq 1$ and it seems I'm stuck.  (And, no, this is not homework!)
 A: Probably easiest:
$$0 \leqslant x \Rightarrow \log (1+x) \leqslant x,$$
so we have (set $x = a_i$ for $1 \leqslant i \leqslant n$)
$$\log \prod_{i=1}^n (1+a_i) = \sum_{i=1}^n \log (1+a_i) \leqslant \sum_{i=1}^n a_i.$$
Further, we have
$$0 \leqslant x \leqslant 1 \Rightarrow e^x \leqslant 1 + 2x,$$
which (set $x = \sum a_i$) together with the above yields the result.


I guess they had a more elementary proof in mind.

I'm not sure if it counts as more elementary, but we can prove it by induction if we prove the stronger inequality
$$\prod_{i=1}^n(1 + a_i) \leqslant 1 + \sum_{i=1}^n a_i + \left(\sum_{i=1}^n a_i\right)^2.$$
The base case $n = 1$ is immediate. For brevity, write $s_n = \sum\limits_{i=1}^n a_i$, then we have
$$\begin{align}
\prod_{i=1}^{n+1}(1 + a_i) &= (1 + a_{n+1})\prod_{i=1}^n (1+a_i)\\
&\leqslant (1 + a_{n+1})(1 + s_n + s_n^2)\\
&= 1 + s_n + a_{n+1} + s_na_{n+1} + s_n^2 + s_n^2a_{n+1}\\
&= 1 + s_{n+1} + s_ns_{n+1} + s_n^2a_{n+1}\\
&\leqslant 1 + s_{n+1} + s_{n+1}(s_n + a_{n+1}s_n)\\
&\leqslant 1 + s_{n+1} + s_{n+1}^2,
\end{align}$$
since by assumption $s_n \leqslant 1$.
A: Here's another proof which is not inductive, and arguably "elementary". The expanded product $\prod_{i=1}^n (1+a_i)$ consists of the initial $1$, plus the sum of the one-fold products, plus the sum of the two-fold products, etc., up to the (single) $n$-fold product. Put $s=\sum_{i=1}^n a_i$ and consider the result of expanding $s^k$, but keeping only those products for which the $a_i$ are distinctly subscripted. Then each such product appears $k!$ times, so that we can bound the sum of the $k$ fold products of the $a_i$  by $s^k/k!.$ Thus we have
$$\prod_{i=1}^n (1+a_i) \le 1 + \frac{s}{1!} + \frac{s^2}{2!}+ \cdots \frac{s^n}{n!}. \tag{1}$$
[This inequality appeared in Hardy's "Course of Pure Mathematics" as an exercise, but seems to have an elementary enough proof.]
So far we have not used $s \le 1$. If we impose it now, we can get an upper bound for the sum on the right side of $(1)$ [with the initial $1$ omitted] by replacing each factorial denominator $t!$ by $2^{t-1}$, and extending the (now geometric) sum to infinity, giving the sum $s/(1-s/2)$, which is at most $2s$ since $s \le 1.$ Thus the right side of $(1)$ is bounded above by $1+2s,$ showing the desired inequality.
