Proof of Bernoulli's inequality The question reads

$$U_n = (1+x)^n - 1 - nx$$
  Show that $U_2 \geq 0$
  Hence or otherwise show that $(1+x)^n \geq 1 + nx$ for all $x \gt -1$.

Obviously the $U_2 \geq 0$ is very easy, I can do that without any trouble but I cannot see how it links to the 2nd part of the question.
 A: I think induction on $n$ would do the trick here, which means you need to show it holds for $n=2$; note that $U_2\geq 0$ is equivalent to showing $(1+x)^2\geq 1+2x$, which holds for all $x$, whether or not they are greater than $-1$. 
Now, assume that $U_n\geq 0$; that is, that $(1+x)^n\geq 1+nx$, and that $x\gt -1$. We want to prove that $U_{n+1}\geq 0$ (i.e., that $(1+x)^{n+1}\geq 1+(n+1)x$).
Take $(1+x)^n\geq 1+nx$. Since $x\gt -1$, then $1+x\gt 0$; multiplying both sides by $1+x$ we get:
$$\begin{align*}
(1+x)^n &\geq 1+nx\\
(1+x)^n(1+x)&\geq (1+nx)(1+x)\\
(1+x)^{n+1}&\geq 1 + nx + x + nx^2\\
(1+x)^{n+1}&\geq 1+ (n+1)x + nx^2 \geq 1+(n+1)x
\end{align*}$$
with the last inequality since $x^2\geq 0$, so $nx^2\geq 0$. 
This proves that if $(1+x)^n\geq 1+nx$ and $x\gt -1$, then $(1+x)^{n+1}\geq 1+(n+1)x$. Since $(1+x)^2\geq 1+2x$, then the result holds for all $n$ by induction.
(This can also be done by calculus, noting that $f(x)=(1+x)^n$ lies above its tangent at $x=0$ on the interval $(-1,\infty)$; the tangent at $0$ is precisely $y=1+nx$.)
A: Instead of proving this, I'll prove a modest generalisation.
Let $x_1,x_2, \ldots x_n$ be numbers greater than $-1$, then $(1+x_1)(1+x_2)\ldots(1+x_n)\geq 1+x_1+x_2+\ldots x_n$
The inductive step is:
$(1+x_1)(1+x_2)\ldots(1+x_{n+1})\geq(1+x_1+x_2+\ldots+x_n)(1+x_{n+1})$
$=\sum_1^{n+1}x_i+x_{n+1}(\sum_1^n x_i)\geq \sum_1^{n+1}x_i$
Bernoulli's inequality follows when all the $x_i$'s are the same.
A: HINT $\rm\ \ U_{n+1}\ =\ (x+1)\ U_n + n\ x^2\ $ which is $\:\ge 0\:$ by induction.
A: My favorite way of proving Bernoulli is to use Jensen inequality. First of all, the inequality is trivial if $1+nx\leq 0$. So suppose that $1+nx>0$.
Following inequality can be proved using Jensen inequality and the fact that $\log$ function is concave:
$$
\frac 1n\log(1+nx)+\frac{n-1}{n}\log 1\leq \log(\frac 1n(1+nx)+\frac{n-1}n)=\log(1+x),
$$
which is the desired inequality. As a matter of fact it does not matter if $n$ is integer here. It suffices that $n\geq 1$ and it is a real number. 
