Proof of equality - possibly previously answered 
I am looking for a proof of the following inequality
  Let a be a positive real number. Prove that for every natural number n
  $$(1+a)^n\ge 1+na.$$

Can you help?
 A: You can prove it by induction :
Base case :
$$1+a \geq 1+a$$
Induction step :
$$(1+a)^{n+1} = (1+a)(1+a)^n \geq (1+a)(1+na) = 1+a+na+na^2 \geq 1+(n+1)a$$
A: Try the binomial theorem.
Then $(1+a)^n = \sum_{k=0}^n \binom{n}{k} a^k$. Taking the first two terms gives
the desired result.
A: It's called Bernoulli's inequality, and you even have more: if $a_1,a_2, \dots, a_n$ are numbers all $>-1$, with the same sign, then:
$$(1+a_1)(1+a_2)\dotsm(1+a_n)\ge 1+a_1+a_2+\dots+a_n.$$
Proof by a simple induction:
\begin{align*}
(1+a_1)(1+a_2)\dotsm(1+a_n)&(1+a_{n+1})\ge (1+a_1+a_2+\dots+a_n)(1+a_{n+1})\\
&=1+a_1+a_2+\dots+a_n+a_{n+1}+ a_{n+1}(a_1+a_2+\dots+a_n)\\
& \ge=1+a_1+a_2+\dots+a_n+a_{n+1}
\end{align*}
since $\,a_{n+1}(a_1+a_2+\dots+a_n)\ge 0\,$ (all numbers have the same sign).
Addendum. In the case $a_1=a_2=\dots=a_n=a$, you have a stronger result: if $a>-1$, then:
$$(1+a)^n\ge 1+na+\frac{n(n-1)}2 a^2.$$
A: Hint $\ \, f_n = (1+a)^n/(1+na)\,$ is a telescoping product of factors $\,\frac{f_{n+1}}{f_n} \ge 1,\,$ so it is $\,\ge 1,\,$ e.g. 
$$\dfrac{(1+a)^4}{1+4a} =\, \dfrac{(1+a)(1+a)}{\qquad\quad\color{#c00}{ 1+2a}}\ \dfrac{(\color{#c00}{1+2a})(1+a)}{\qquad\quad\ \ \color{#0a0}{1+3a}}\ \dfrac{(\color{#0a0}{1+3a})(1+a)}{\qquad\quad\ \ 1+4a}\qquad $$
Each factor is $> 1$ by $\ \underbrace{(1 + na)(1 + a)}_{\large\rm numerator}-\,\underbrace{(1+(n\!+\!1)a)}_{\large\rm denominator} =\, na^2 \ge\, 0 $
Remark $\ $ Equivalently $\, f_n\,$ is increasing $f_{n+1}\ge f_n\,$ so $\ f_n\ge f_1 = 1\,$  by induction. 
