Binomial Inequality: $\forall x \in \mathbb{R}, x\geq-1, \forall n \in \mathbb{N}: (1+x)^{n} \geq 1+nx$
I need to prove this using the Mean Value Theorem: If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, $\exists$ $c \in (a,b)$:
$\dfrac{f(b)-f(a)}{b-a} = f'(c)$
If I set $f(x)=(1+x)^n$, I get $n(1+a)^{n-1}(b-a) \lt (1+b)^n – (1+a)^n \lt n(1+b)^n (b-a)$ using the fact that if $f(x)$ is increasing, $a \lt c \lt b \implies f'(a) \lt f'(c) \lt f'(b)$ and then substituting the earlier exppression for $f'(c)$.
I'm not really sure what to do next.
Best Answer
For $x\gt0$ the mean value theorem on $f(x)=(1+x)^n$ on $[0,x]$ gives $\dfrac{(1+x)^n-1}{x}=n(1+c)^{n-1}$ for some $c\in(0,x)$.
Therefore $\dfrac{(1+x)^n-1}{x}=n(1+c)^{n-1}\gt n$ and the result follows.
The case $-1\leq x\lt0$ is similar. The case $x=0$ is simpler.