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?

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    $\begingroup$ Not relevant to your question but $\frac{x-1}{2} \leq \log(1+x)$ is not sufficient to prove the claim, you need something like $\frac{x}{2} \leq \log(1+x)$ (which is also true in the relevant interval). $\endgroup$ Commented Sep 12, 2022 at 9:48
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    $\begingroup$ @AsafKaragila May I ask why you removed this question from network questions? You did not leave any comment when you took that significant action. $\endgroup$
    – Apass.Jack
    Commented Sep 12, 2022 at 21:55
  • $\begingroup$ Under what criteria should we remove questions from the HNQ list?. $\endgroup$
    – Apass.Jack
    Commented Sep 12, 2022 at 22:41
  • $\begingroup$ @Apass.Jack: Unfortunately, the removal from the HNQ is a non-transparent process. It is mainly done by one moderator who has made it clear that he would gladly “remove any and all questions from the HNQ list.” I had started some discussion in the Math Meta Chat some time ago, but all I got is that “we do not want bad questions on the HNQ.” I have seen good questions removed, and bad questions preserved. $\endgroup$
    – Martin R
    Commented Sep 13, 2022 at 7:29

2 Answers 2


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

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 $0\le x\le1$, $y\ge0$.

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


Let $x_i,y_i>0$ such that :


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 :


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 :


Then :

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

  • $\begingroup$ To give two equality cases we can choose for : $$-\left(n+ax-n\left(1+2x\right)^{\frac{1}{n}}\right)\geq 0$$ as $a=\left(3^{\frac{1}{n}}-1\right)n$ for $0\leq x\leq 1$ and $n\geq 1$ $\endgroup$ Commented Sep 12, 2022 at 14:42
  • $\begingroup$ Is this a “a more elementary proof”? $\endgroup$
    – Martin R
    Commented Sep 13, 2022 at 8:14
  • $\begingroup$ Btw, $n+x\leq n(1+2x)^{\frac{1}{n}}$ follows from $$\left(1+\frac xn\right)^n \le e^x \le 1+(e-1)x \le 1+2x$$ where the middle inequality holds for $x \in [0, 1]$ because $\exp$ is convex. $\endgroup$
    – Martin R
    Commented Sep 13, 2022 at 8:33
  • $\begingroup$ @MartinR It's an elementary proof you can believe it and you forgot we cannot use exponential . $\endgroup$ Commented Sep 13, 2022 at 8:36

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