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Given $a_1,a_2,\dots,a_n>0$ and $\mathbb{S}=a_1+\dots+a_n<1$ Prove: $$ 1+\mathbb{S}<(1+a_1) \dot\ \dots \dot\ (1+a_n)<\frac{1}{1-\mathbb{S}} $$

I proved the left inequality, but I can't figure the right one.

Any help appreciated.

S.C.B.
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Itay4
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2 Answers2

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Let $S=\sum_{k=1}^n a_k<1$, where $a_k>0$ for all $k$.

METHODOLOGY $1$: Use the AM-GM Inequality

Using the AM-GM Inequality revelas

$$\begin{align} \prod_{k=1}^n(1+a_k)&\le \left(\frac{\sum_{k=1}^n(1+a_k))}{n}\right)^n\\\\ &=\left(1+\frac{S}{n}\right)^n\tag 1\\\\ \end{align}$$

Now, in THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's inequality that (i) the sequence $\left(1+\frac xn\right)^n$ monotonically increases to $e^x$ and (ii) that the exponential function satisfies the inequality $e^x\le \frac{1}{1-x}$ for $x<1$.

Therefore, we see that

$$\begin{align} \prod_{k=1}^n(1+a_k)&\le e^S\\\\ &\le \frac{1}{1-S} \end{align}$$

as was to be shown!

METHODOLOGY $2$: Use a Standard Inequality for the Loagarithm Function

Again, in THIS ANSWER, I showed that the logarithm satisfies the inequality $\log(x)\le x-1$.

Hence, we have

$$\begin{align} \log\left(\prod_{k=1}^n(1+a_k)\right)&=\sum_{k=1}^n\log(1+a_k)\\\\ &\le \sum_{k=1}^na_k\\\\ &=S \end{align}$$

Therefore, we see that

$$\prod_{k=1}^n(1+a_k)\le e^S\le \frac{1}{1-S}$$

as expected!

Community
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Mark Viola
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  • The transition from $(2)$ to $(3)$ is proven only for $x<1$. Could you please explain that? – Itay4 Mar 06 '17 at 15:23
  • @Itay4 I've edited to provide a reference to an answer I posted which develops the inequality of interest using only elementary methods. – Mark Viola Mar 06 '17 at 15:30
  • Nice answer. Alternatively, we can write $\prod_{k = 1}^n (1 + a_k) < \prod_{k = 1}^n e^{a_k} = e^S < \frac{1}{1 - S}$. – kobe Mar 06 '17 at 15:32
  • @kobe Thank you! Yes, that is effectively equivalent to the second approach, albeit more direct. – Mark Viola Mar 06 '17 at 15:33
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As you proved the left one yourself, here is the proof of the right ineq. by induction:

$$(1+a_1)(1-a_1)=1-a_1^2<1 \Rightarrow 1+a_1<\frac{1}{1-a_1}$$

Suppose for some $n-1\in \Bbb N$, $$(1+a_1)...(1+a_{n-1})<\frac{1}{1-\Bbb S_{n-1}}$$

Then if $a_1,...,a_n>0$ and $\Bbb S<1$, we have $$(1+a_1)...(1+a_n)<\frac{1+a_n}{1-\Bbb S_{n-1}}=\frac{1+a_n}{1-\Bbb S_{n-1}}\frac{1-a_n}{1-a_n}=\frac{1-a_n^2}{1-\Bbb S_n +a_n\Bbb S_{n-1}}<\frac{1}{1-\Bbb S_n}$$ where $\Bbb S_{j}=\sum_{i=1}^{j}a_i$

Result follows

Nick
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  • Nice and straightforward. I find producing this kind of proof just involves using the induction hypothesis and plugging onwards. – marty cohen Mar 08 '17 at 03:45