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I want to prove $$\prod_{k=1}^n(1+a_k)\leq1+2\sum_{k=1}^n a_k$$ if $\sum_{k=1}^n a_k\leq1$ and $a_k\in[0,+\infty)$

I have no idea where to start, any advice would be greatly appreciated!

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  • $\begingroup$ Have you tried induction? Seems like the most natural and logical way to start. $\endgroup$ – windircurse May 5 '16 at 16:15
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    $\begingroup$ @windircursed Have you tried induction? The obvious induction argument doesn't quite work when I try it. $\endgroup$ – David C. Ullrich May 5 '16 at 16:31
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The simple induction would work with no problem if we were given that $\sum a_k\le 1/2$. Which is enough for the presumed application to infinite products, but doesn't solve the current problem.

Hint for the problem as given: $$\log(1+t)\le t\quad(t\ge0)$$ and $$\log(1+2t)\ge t\quad(0\le t\le 1).$$

You can prove both these by writing the logarithm as the integral of $ds/s$. Oops, no you can't. There was a bit of nonsense in the proof I had in mind.

The first inequality is clear; if $t>0$ then $$\log(1+t)=\int_1^{1+t}\frac{ds}s\le\int_1^{1+t}\,ds=t.$$ The second is slightly more subtle. The second derivative of $\log(1+t)$ is negative; hence $\log(1+t)/t$ is a decreasing function of $t>0$, so $$\frac{\log(1+2t)}{t}\ge\log(3)>1\quad(0<t\le 1).$$The same argument shows that in fact $$\log(1+(e-1)t)\ge t\quad(0\le t\le 1),$$giving the sharper inequality Winther noted: $$\log\left(\prod(1+a_k)\right)=\sum\log(1+a_k)\le\sum a_k\le\log\left(1+(e-1)\sum a_k\right).$$

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  • $\begingroup$ Oh really? What's the problem? $\endgroup$ – David C. Ullrich May 5 '16 at 16:46
  • $\begingroup$ I'm not quite sure how to use your hints. I tried induction but now I'm stuck at $\prod_{k=1}^n (1+a_k)\leq 2$ $\endgroup$ – ntm May 5 '16 at 16:48
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    $\begingroup$ @ntm Induction doesn't work! I said so! (Ok, who knows whether there's some inductive argument that works, but the obvious induction simply does not work.) Take the logarithm of both sides of the inequality and you should see the relevance of the hints. You'll also need the fact that the logarithm of a product is the sum of the logarithms... $\endgroup$ – David C. Ullrich May 5 '16 at 16:53
  • $\begingroup$ oh... so it leads to this: $log(\prod_{k=1}^n (1+a_k))=\sum_{k=1}^n log(1+a_k)\leq \sum_{k=1}^n a_k\leq log(1+2\sum_{k=1}^n a_k)$ $\endgroup$ – ntm May 5 '16 at 17:04
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    $\begingroup$ @ntm Right. This is where you click on the little arrow and the little checkmark, btw... $\endgroup$ – David C. Ullrich May 5 '16 at 17:06
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David C. Ullrich's answer shows a very simple way to do this. Here is an altenative which is more complicated but which can allows us to derive a sharper inequality if needed. We treat the problem as an optimization problem of maximizing the product $P = \prod_{k=1}^n(1+a_k)$ over $a_k\geq 0$ under the constraint $S = \sum_{k=1}^n a_k = r$ for some $0\leq r \leq 1$.

We put up the Lagrangian $$L = P - \lambda(r-S)$$

The extemal point satisfies

$$\frac{\partial L}{\partial a_i} = \frac{P}{1+a_i} - \lambda \implies 1+a_i = \frac{P}{\lambda} \implies a_1=a_2=\ldots = \frac{r}{n}$$ This is a local maximum and we can rule out the possibillity of any maximum points lying on the boundary $a_i = 0$ so the point is a global maximum. This shows that

$$\max_{\matrix{\{a_i\}_{i=1}^n\in[0,\infty)\\\sum a_k = r}}\prod_{k=1}^n(1+a_k) = \left(1+\frac{r}{n}\right)^n \leq e^r$$

so the problem reduces to showing that $1+2r-e^r\geq 0$ for $r\in[0,1]$ which follows by, say, using Taylors theorem.


By the same approach we can deduce the slightly stronger result

$$\prod_{k=1}^n(1+a_k) \leq 1 + \frac{e^r-1}{r}\sum_{k=1}^n a_k~~~\text{when}~~~\sum_{k=1}^n a_k \leq r \leq 1$$ in particular for $r=1$ which is the problem at hand we can reduce the constant $2$ down to $e-1 \simeq 1.71$.

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  • $\begingroup$ What I did gives $e-1$ just as well. Otoh I'm glad you mentioned this; thinking about it I noticed that the argument I had for the main inequality needed to be fixed. $\endgroup$ – David C. Ullrich May 6 '16 at 19:44

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