If $a_1,\ldots ,a_n >0$ and $S=a_1+\ldots + a_n <1$, then $(1+a_1)(1+a_2)\ldots (1+a_n)(1-S) < 1$

Hint: The right hand side is the arithmetic mean of $1 + a_1, \ldots, 1 + a_n$ and $1 - S$.

Note that $$(1+x_1)(1+x_2)\ldots(1+x_n)=1+(x_1+x_2+\ldots+x_n)+(x_1x_2+x_1x_3+\ldots+x_{n-1}x_n) + (x_1x_2x_3+x_1x_2x_4+\ldots) + \ldots + (x_1x_2x_3\ldots x_{n-1}+\ldots)+x_1x_2\ldots x_n.$$

Now, \begin{align*} 1&=1\\ x_1+x_2+\ldots+x_n &= S \\ x_1x_2+x_1x_3+\ldots+x_{n-1}x_n &< S^2 \\ x_1x_2x_3+x_1x_2x_4+\ldots &< S^3 \\ &\vdots \\ x_1x_2x_3\ldots x_{n-1}+\ldots &< S^{n-1}\\ x_1x_2\ldots x_n &< S^n. \end{align*}

It follows that $$(1+x_1)(1+x_2)\ldots(1+x_n) < 1+S+S^2+\ldots+S^n = \frac{1-S^{n+1}}{1-S} < \frac{1}{1-S},$$ as desired.

Hint: use induction. One quick way of going from $n$ to $n+1$ is to differentiate w.r.t. $a_{n+1}$. You will quickly see that the derivative is negative; since the limiting value as $a_{n+1} \to 0$ is less than one by induction hypothesis the result follows.