Use induction to show $(1+a)^n \ge na$ for $n\in \mathbb{N}$ and $a>0$. I am familiar with Bernoulli's inequality which is quite straightforward to prove using induction, but this problem (although simpler at first glance) seems to be more complicated.  What do you guys think?
Please stop posting solutions on proving Bernoulli's inequality.  I am interested in using only induction to solve the stated problem.
Can anyone solve this problem directly using only induction?
Yes I am aware the inequality is actually strict, but I have just copied the problem directly from a high school textbook.
I am not interested in just a solution, but more interested in how would a high school student who just learned about induction solve this problem?
 A: First of all I would like to tell that the equality will never hold true. So the real question is to prove $$(1+a)^n>na$$ by induction. Now coming to the question, we can see that at $n=1$ the inequality is true. Assume that, at $n=n$ the inequality is true. Now let's see at $n=n+1$
RHS$=na+a$ it will be maximum when both $na$ and $a$ are maximum. Since, $a$ is a constant it's value depends entirely on $na$. This will be maximum when it is approaching to $(1+a)^n$ For a time being let's assume it's value as $(1+a)^n$
So, $RHS_{max}=(1+a)^n+a$
LHS$=(1+a)^{n+1}=(1+a)^n(1+a)$
Let $(1+a)^n=k$
So, $RHS_{max}=k+a$ and LHS$=(1+a)^{n+1}=k(1+a)=k+ka$
RHS=LHS iff $ka=a$ which will only happen if $k=1$ or $a=0$ But $a$ cannot be $0$ as it is given that $a>0$ So $k$ has to be $1$ which means that $(1+a)^n$ has to be $1$ which can happen if $n=0$ or $a=0$.
Again $n \in \mathbb{N}$ and $a>0$ therefore it can't happen.
$\implies$ $RHS_{max}<LHS$ or $(1+a)^n+a<(1+a)^n(1+a)$ or $$(n+1)a<(1+a)^{n+1}$$
So, we see that even if RHS is equal to $(1+a)^n+a$ it is less than LHS. Since actually RHS $<(1+a)^n+a$ so it means that
$$RHS <(1+a)^n+a<(1+a)^{n+1}=LHS$$
A: Edit:This is the proof of Bernoulli's inequality. I don’t really answered the real question.
Here's my approach. Let’s prove a stronger proposition, namely, $(1+a)^n\ge 1+na$ for $n\in\mathbb{N}$ and $a>0.$
For the base case $n=1$, the proof is trivial.
Suppose, that $\forall a>0((1+a)^k\ge 1+ka)$ for some integer $k>1$. Now we claim that $\forall a>0((1+a)^{k+1}\ge 1+(k+1)a)$. Since $(1+a)^{k+1}=(1+a)^k(1+a)\ge (1+ka)(1+a)\ge1+(k+1)a$ for all $a>0$, it follow that our claim is true, this closed the induction.
A: Here is my solution.
First note for $n=1$ we have $$(1+a)^1=1+a \ge a=(1)a$$
We now assume $\exists k \in \mathbb{N}$ such that $$(1+a)^k\ge ka \hspace{10mm}(*)$$
Finally, consider $$\begin{eqnarray}(1+a)^{k+1} &=&(1+a)(1+a)^k \\ &=&(1+a)^k+a(1+a)^k \hspace{10mm} \text{Distribute $(1+a)^k$}\\ &\geq& (1+a)^k+a \hspace{27mm} \text{Since } (1+a)^k>1 \\ &\ge & ka +a \hspace{38.4mm}\text{By }(*) \\ &=&(k+1)a   \end{eqnarray}$$
Hence, by the principle of mathematical induction we can conclude $(1+a)^n\ge na$ for all $n\in \mathbb{N}$.
