# Thread: simple proof by induction question

1. ## simple proof by induction question

I have come across this 'simple' question in more than one book on proofs, yet I am not able to find a solution:

Prove, for all $\displaystyle n\geq0$, that $\displaystyle (1+x)^n\geq1+nx$ if $\displaystyle 1+x>0$.

I can see how this fact can be demonstrated via binomial theorem

$\displaystyle (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$

(since the first third component of the expansion is always non-negative)

but could you give me a hint how to go about proving it using proof by induction?

thanks

2. Originally Posted by Volga
I have come across this 'simple' question in more than one book on proofs, yet I am not able to find a solution:

Prove, for all $\displaystyle n\geq0$, that $\displaystyle (1+x)^n\geq1+nx$ if $\displaystyle 1+x>0$.

I can see how this fact can be demonstrated via binomial theorem

$\displaystyle (1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...$

(since the first third component of the expansion is always non-negative)

but could you give me a hint how to go about proving it using proof by induction?

thanks
Merely note that $\displaystyle \left(1+x)^{n+1}=(1+x)^n(1+x)\geqslant (1+nx)(1+x)=1+nx+x+nx^2=1+(n+1)x+nx^2\geqslant 1+(n+1)x$.

3. I would try

For n=1: $\displaystyle \displaystyle (1+x)^1 \geq 1+x$ True!

Assume n=k: $\displaystyle \displaystyle (1+x)^k \geq 1+kx$ is also true.

Now for n=k+1:

$\displaystyle \displaystyle (1+x)^{k+1} \geq 1+(k+1)x$

$\displaystyle \displaystyle (1+x)^{k}(1+x)^{1} \geq 1+kx+x$

Now use $\displaystyle \displaystyle (1+x)^k \geq 1+kx$ to finish the argument.

4. Oh thanks! I was stupid enough to use x=k and x=k+1 instead of n=k and n=k+1!!! Now it is obvious...