Prove $\prod_{k=1}^n(1+a_k)\leq1+2\sum_{k=1}^n a_k$

Question

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!

Answer 1

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: $$\log\left(\prod(1+a_k)\right)=\sum\log(1+a_k)\le\sum a_k\le\log\left(1+(e-1)\sum a_k\right).$$

Answer 2

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.

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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 $e-1 \simeq 1.71$.