(Bernoulli's inequality;Jac.Bernoulli 1689,see 1744,Opera,p.380;Barrow 1670,see 1860,Works,Lectio VII,XIII,p.224).By induction on $n$,prove that
1.\begin{equation}
\label{eq:1.18}
(1+a)^n\geq 1+na ~~ \mbox{for}~~ a\geq -1,n=0,1,2,\cdots
\end{equation}
Proof:When $n=0$,$1=1$.Suppose when $n=k(k\geq 0)$,the inequality also hold.That is
\begin{equation}
\label{eq:1.39}
(1+a)^k\geq 1+ka
\end{equation}
Then
\begin{equation}
\label{eq:1.41}
(1+a)^{k+1}=(1+a)^k(1+a)\geq (1+ka)(1+a)=1+ka+a+ka^2\geq 1+(k+1)a
\end{equation}
Done.
2.\begin{equation}
\label{eq:1.55}
1-na<(1-a)^n<\frac{1}{1+na}~~\mbox{for}~~0<a<1,n=2,3,\cdots
\end{equation}
Proof:
\begin{equation}
\label{eq:2.44}
\frac{1-(1-a)^n}{a}<n
\end{equation}
Let $f(x)=x^n$.According to the differential mean value theorem,we just need to prove that when $0<x<1$,
\begin{equation}
f'(x)=nx^{n-1}<n
\end{equation}
This is obvious.
Now I prove
\begin{equation}
(1-a)^n<\frac{1}{1+na}
\end{equation}
Let $a=\frac{1}{k}$.We just need to prove that
\begin{equation}
(1+\frac{1}{k-1})^n>1+\frac{n}{k}
\end{equation}
This is true because
\begin{equation}
\label{eq:'}
(1+\frac{1}{k-1})^n\geq 1+\frac{n}{k-1}>1+\frac{n}{k}
\end{equation}