# Multiplicative sum is bigger than sum prove

I want to prove this: $$\prod_{i=1}^n(1+a_i) \ge 1+\sum_{i=1}^n(a_i)$$

Whereas $a_n$ is a family of non-negative, real numbers.

I wrote out the sums and got to this point: $$(1+a_1)*(1+a_2)* \ ... \ *(1+a_n) \geq (a_1+\frac1n) + (a_2+\frac1n) + ...+(a_n + \frac1n)$$

And I know that every part of the sum is now for sure smaller than every factor of the multiplicative sum. However, this does not suffice to prove the statement, since for example $1,05 * 1,04 < 1,03+1,02$ .

How do I finish the proof?

• Induction works.
– user339727
Commented Nov 7, 2016 at 10:20

If you explicitly multiply the LHS you will get $$\prod_{i=1}^n(1+a_i)=1+\sum_{i=1}^n a_i+\text{many other nonnegative terms} \ge 1+\sum_{i=1}^n a_i.$$

You can use a neat induction on $n$.

First let us show that your statement is trivially true for $n=2$.

$$\prod_{i=1}^2 (1+a_i) \ge 1 + \sum_{i=1}^2 a_i \iff$$

$$\iff (1 + a_1)(1 + a_2) \ge 1 + a_1 + a_2$$

which is obvious if you factor out the left-hand side.

Now assuming that your inequality is true for $n$, let us show it is also true for $n+1$:

$$\prod_{i=1}^{n+1} (1+a_i) \ge 1 + \sum_{i=1}^{n+1} a_i \iff$$

$$\iff (1 + a_{n+1})\prod_{i=1}^n (1+a_i) \ge 1 + a_{n+1} + \sum_{i=1}^n a_i$$

Again expanding the left hand side should now make it obvious:

$$(1 + a_{n+1})\prod_{i=1}^n (1+a_i) \ge 1 + a_{n+1} + \sum_{i=1}^n a_i \iff$$

$$\iff \prod_{i=1}^n (1+a_i) + a_{n+1}\prod_{i=1}^n (1+a_i) \ge a_{n+1} + (1 + \sum_{i=1}^n a_i) \iff$$

but you know, from the inductive step, that

$$\prod_{i=1}^{n} (1+a_i) \ge 1 + \sum_{i=1}^{n} a_i$$

so it suffices to show that

$$a_{n+1}\prod_{i=1}^n (1+a_i) \ge a_{n+1}$$ but that is trivially true since

$k = \prod_{i=1}^n (1+a_i) \ge 1$; then you can write $ka_n \ge a_n$ which holds true in our case: $a_n \ge 0$ and $k \ge 1$.

• Thanks! Just a question: How should the final expansion make it obvious? I mean, it is obvious in my head, but on paper I do not know how to formulate it correctly. Commented Nov 7, 2016 at 10:59
• Fantastic! Thanks a lot! Commented Nov 7, 2016 at 11:09
• @M3xr I edited the conclusion so that you can see the steps needed.
– RGS
Commented Nov 7, 2016 at 11:10

You can show the inequality by using induction. For $n=1$ we do not have to show anything. Assume that the inequality holds for $n\in \mathbf N$. Then we get $$\prod_{i=1}^{n+1}(1+a_i)=(1+a_{n+1})\prod_{i=1}^{n}(1+a_i) \stackrel{\text{IH}}{\geq} (1+a_{n+1}) \left(1+\sum_{i=1}^n a_j \right) =1+ a_{n+1} \sum_{i=1}^n a_j + a_{n+1} +\sum_{i=1}^n a_j.$$ By assumption $a_{n+1} \sum_{i=1}^n a_j >0$ and therefore we get $$1+ a_{n+1} \sum_{i=1}^n a_j + a_{n+1} \sum_{i=1}^n a_j > 1+ a_{n+1} \sum_{i=1}^n a_j = 1+ \sum_{i=1}^{n+1} a_j.$$ By the way: the inequality $$\prod_{i=1}^{n}(1+a_i) > 1+ \sum_{i=1}^{n} a_j$$ also holds if $a_i \in (-1,0)$ for all $i\in \{1, \dotsc, n\}$ and if $a_i=a$ for all $i\in \{1, \dotsc, n\}$ you get the well known Bernoulli's inequality.

• Thank you very much. I am just not sure whether you made a typo or not in the end of the first line: Shouldn't it be: $1+ a_{n+1} * \sum_{i=1}^na_i + a_{n+1} + \sum_{i=1}^na_i$ ?? So instead of a $*$ , shouldn't there be a $+$ between $a_{n+1}$ and the sum? Commented Nov 7, 2016 at 10:57
• Yes, this is a typo. Thank you.
– user339727
Commented Nov 7, 2016 at 11:07