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

Let $$a_1,\ldots ,a_n >0$$ and $$S=a_1+\ldots + a_n <1$$.

I want to show that: $$(1+a_1)(1+a_2)\ldots (1+a_n)(1-S) < 1$$

So by expanding the LHS I get: $$1+S-S+a_1\ldots a_n + a_1\ldots a_{n-1} +\ldots + a_1 a_2+\ldots$$

I want to show somehow that the LHS equals: $$1-\ldots$$ where $$\ldots >0$$, but I dont see how exactly, can anyone help with this?

Thanks!

• Hello, I think you can prove it by induction on the number of variables, $n$. Jul 27, 2019 at 11:57
• Surely there's a nicer proof than induction, right? Jul 27, 2019 at 11:59
• Are you sure your expansion is correct? If all the $a_i$ are positive, your expression would be greater than $1$. Jul 27, 2019 at 12:01
• So, you want the product of the $a_i$s to be positive, not each individual $a_i$? What if $n = 2$, $a_1 = a_2 = -3$? Jul 27, 2019 at 12:03
• @TheoBendit yeah, that was a question of my student; I assume there's a mistake in the question as you noticed. In that case it should be for each $a_i$. I'll edit it. Jul 27, 2019 at 12:07

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.