Elementary proof of "generalized reverse Bernoulli inequality"

Question

I've stumbled upon the following exercise in an early chapter of an analysis textbook:

Let $a_n$ be a finite, nonnegative sequence such that $\sum_{i=0}^n a_i\le 1$. Prove $$ \prod_{i=1}^n (1 + a_i) \le 1 + 2 \sum_{i=1}^n a_i. $$

There seems to be a relatively simple proof using the exponential function utilizing the fact that for $0 \le x \le 2$, $\frac{x - 1}{2} ≤ \log(1 + x) ≤ x$, as remarked by a friend on twitter.

However, this exercise appears in the textbook before $e^x$ or even $3^x$ are defined for real $x$, so I'm wondering whether there is a more elementary proof. A simple induction-based method certainly does not work, as it leaves you with a target statement that doesn't even hold necessarily. So I'm wondering, is anyone aware of a proof of this not relying on exponentials?

Answer 1

A powerful technique is to guess a stronger proposition that will enable induction as well.

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Let $a_i\ge0$, $\sum_{i=1}^n a_i\le 1$. We have the following stronger inequality.

$$ \prod_{i=1}^n (1 + a_i) \le1+(1+\sum_{i=1}^na_i)\sum_{i=1}^n a_i. $$

Proof: $n=1$ is trivial.

The induction step is guaranteed by the following inequality that holds for $0\le x\le1$, $y\ge0$.

$$(1+x+x^2)(1+y)\le 1+ (x+y)+(x+y)^2.$$

Answer 2

Let $x_i,y_i>0$ such that :

$$\sum_{i=1}^{n}x_i\geq\sum_{i=1}^{n}y_i$$

And :

$$|x_i-x_j|\leq |y_i-y_j|,1\leq i\leq n,1\leq j\leq n,i\neq j$$

Then we have :

$$\prod_{i=1}^{n}x_i\geq\prod_{i=1}^{n}y_i$$

Here $x_i=\left(1+2\sum_{i=1}^{n}a_i\right)^{\frac{1}{n}},y_i=1+a_i$

We need to show that :

$$n+\sum_{i=1}^{n}a_i\leq n\left(1+2\sum_{i=1}^{n}a_i\right)^{\frac{1}{n}}$$

Or $x=\sum_{i=1}^{n}a_i$:

$$n+x\leq n(1+2x)^{\frac{1}{n}}\tag{I}$$

Using a tricky Bernoulli's inequality we have for $0<x\leq 0.5$:

$$n+x-\frac{n\left(1+\left(1+\frac{1}{n}\right)2x\right)}{1+2x}\geq n+x-n(1+2x)^{\frac{1}{n}}$$

For the case $0.5<x\leq 1$ you can use a chord as we have :

$$f(x)=n(1+2x)^{\frac{1}{n}}-n-x\geq \frac{\left(f\left(0.5\right)-f\left(1\right)\right)}{0.5-1}\left(x-0.5\right)+f\left(0.5\right)$$

Because $f(x)$ is concave .

Nota Bene :

Let :

$$h\left(x\right)=n+x-n\left(1+2x\right)^{\frac{1}{n}}$$

Then :

$$\lim_{n\to \infty}h'(0.5)=0$$