Interpretation of a statement that I think I can prove by mathematical induction

Question

I want to prove by mathematical induction the following statement: for any sequence (a_k) of non negative real numbers holds that:

$$(1+a1)(1+a2)...(1+a_n) >= 1 + a_1 + a_2 + ... a_n$$

I think this statement verbally translates as: the products of the sum of any positive real number with 1 is greater than or equal to the sum between 1 and the positive real numbers chosen.

Indeed, I have a doubt now. Do the indices of the letters $a$ tell me that inside any parenthesis can be or should be chosen a different real number?.

If yes, I would be a bit in trouble in writing the base case - or might I skip it, in some circumstances? - otherwise, since that we now deal with positive real numbers - a superset with respect to the natural numbers - I think that I could choose zero for the base case - perhaps, I should choose zero - as it is the first non negative real number! Obviously, you can disprove my statement.

However, zero holds for the base case, because we would end up with 1 = 1 - but I am not considering the different indices of the letters $a$, which is my first doubt.

Answer 1

The base case is $n=1$ because your first available subscript is $1$. There is no $a_0$. So base case, $n=1$ you must show that $(1+a_1) \geq 1 + a_1$, which is trivial. Then prove induction statement for $n+1$, which states \begin{equation} \prod_{i=1}^{n+1} (1+a_i) \geq 1 + \sum_{i=1}^{n+1} a_i \end{equation} But \begin{equation} \prod_{i=1}^{n+1} (1+a_i) = \prod_{i=1}^{n} (1+a_i) \cdot (1+a_{n+1}) \end{equation} And \begin{equation} 1 + \sum_{i=1}^{n+1} a_i = 1 + \sum_{i=1}^{n} a_i + a_{n+1} \end{equation}

Consider $\prod_{i=1}^{n} (1+a_i) \cdot (1+a_{n+1})$. Rewrite using distributive law such that \begin{equation} \prod_{i=1}^{n} (1+a_i) \cdot (1+a_{n+1})=\prod_{i=1}^{n} (1+a_i) + a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \end{equation}

Since you are using mathematical induction, you are assuming that your induction statement \begin{equation} P(n):\prod_{i=1}^{n} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i \end{equation} is true. Now add $a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i)$ to both sides of the inequality such that \begin{equation} \prod_{i=1}^{n} (1+a_i) +a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i + a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \end{equation}

The left-hand side of this inequality is just \begin{equation} \prod_{i=1}^{n+1} (1+a_i) \end{equation}

as was shown near the start. Upon substitution into the inequality,

\begin{equation} \prod_{i=1}^{n+1} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i + a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \end{equation}

Now, since every $a_i$ is a nonnegative integer, $1+a_i\geq 1$. More specifically,

\begin{equation} \prod_{i=1}^{n} (1+a_i) \geq 1 \implies a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \geq a_{n+1} \end{equation}

Thus,

\begin{equation} \prod_{i=1}^{n+1} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i + a_{n+1} \cdot \prod_{i=1}^{n} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i + a_{n+1} \\ \prod_{i=1}^{n+1} (1+a_i) \geq 1 + \sum_{i=1}^{n} a_i + a_{n+1} =1 + \sum_{i=1}^{n+1} a_i \\ \prod_{i=1}^{n+1} (1+a_i) \geq 1 + \sum_{i=1}^{n+1} a_i \end{equation}

Done

Answer 2

Consider a sequence $a_n \geq 0$. Clearly $$a_1+1 \geq a_1+1.$$ Assume that for $k$ we have $$(1+a_1)(1+a_2) \cdots (1+a_k) \geq 1+a_1+a_2 + \cdots +a_k.$$ Now $$(1+a_1)(1+a_2) \cdots (1+a_k)(1+a_{k+1}) \geq \left( 1+a_1+a_2 + \cdots +a_k \right)(1+a_{k+1}) \\ \geq 1+a_1+a_2 + \cdots +a_k +a_{k+1}, $$ because the last expression is obtained by dropping some non-negative terms.