1. ## Induction

The question I have is....
Using induction, verify the inequality

$\displaystyle (1+x)^n$ is greater than or equal to (1+nx)
for x is greater than or equal to -1 and n is greater than or equal to 1

The basis step was easy enough (as always)
However, I became stuck at the induction part...
Since the condition given is for BOTH x AND n, I tried doing induction for
x+1 and n+1

Assuming the above statement to be true, prove
(2+x)^(n+1) is greater than or equal to 1+(n+1)(x+1)

I tried a lot of different ways, but couldn't solve it...
The closest thing I got to an answer was by doing the following...

(2+x)^(n+1)=(2+x)(2+x)^n

(2+x)(2+x)^n is greater than or equal to (1+x)^n+(n+x+1)

Since, (2+x)^n is greater than or equal to (1+x)^n I assumed that part of the equation to be true... leaving me to prove
(2+x) is greater than or equal to (n+x+1)/((1+x)^n)

which is easy enough to prove.... Since $\displaystyle (1+x)^n$ will ALWAYS be greater than 1 given the condition x is greater than or equal to -1 and n is greater than or equal to 1

But I'm pretty sure this is NOT the right way to prove this (since this is supposed to be proven using induction... and I feel like I am taking way too many assumptions)

Any hints/tips on how to solve an induction problem with two changing variables?

2. ## Re: Induction

Using induction:
First step
$\displaystyle n=1$ then $\displaystyle (1+x)\geq 1+x$ which is true.
Second step
Suppose the proposition is true for $\displaystyle n=k$ and so:
$\displaystyle (1+x)^{k}\geq 1+kx$
We have to prove the proposition is true for $\displaystyle n=k+1$ and so for all $\displaystyle n\geq 1$ (by the given condition).
Third step
Proof for $\displaystyle n=k+1$ that means:
$\displaystyle (1+x)^{k+1}\geq 1+(k+1)x$
We now for sure that:
$\displaystyle (1+x)^{k}\cdot (1+x)\geq (1+kx)\cdot (1+x)$ (if $\displaystyle (1+x)>0$)
(This is true by step 2)
That means:
$\displaystyle (1+x)^{k}\cdot (1+x)\geq 1+kx+x+kx^2$
And because $\displaystyle x^2>0$: $\displaystyle 1+kx+x+kx^2\geq 1+kx+x$, therefore:
$\displaystyle (1+x)^{k}\cdot (1+x)\geq 1+kx+x+kx^2\geq 1+kx+x$
$\displaystyle (1+x)^{k+1}\geq 1+kx+x$
$\displaystyle (1+x)^{k+1}\geq 1+(k+1)x$

Which is true by step 2.

3. ## Re: Induction

Originally Posted by HyunGyum
The question I have is....
Using induction, verify the inequality

$\displaystyle (1+x)^n$ is greater than or equal to (1+nx)
for x is greater than or equal to -1 and n is greater than or equal to 1

The basis step was easy enough (as always)
However, I became stuck at the induction part...
Since the condition given is for BOTH x AND n, I tried doing induction for
x+1 and n+1

Assuming the above statement to be true, prove
(2+x)^(n+1) is greater than or equal to 1+(n+1)(x+1)

I tried a lot of different ways, but couldn't solve it...

Any hints/tips on how to solve an induction problem with two changing variables?
$\displaystyle x\ge\ -1$

so that

$\displaystyle 1+x\ge\ 0\Rightarrow\ (1+x)^n\ge\ 0$

Hence we are dealing with non-negative values for that part.

That is the significance of the constraint on x.

To apply Proof By Induction, continue in the normal way using n=k and n=k+1.

You want to prove that IF $\displaystyle (1+x)^k\ge\ 1+kx$

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

Therefore

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

So if $\displaystyle (1+x)^k\ge\ 1+kx$

then $\displaystyle (1+x)(1+x)^k\ge\ (1+x)(1+kx)$

Now you only need show that $\displaystyle (1+x)(1+kx)\ge\ 1+(k+1)x$

when k is positive (since k is any value of n).