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#### skatenerd

##### Active member

- Oct 3, 2012

- 114

I started by supposing \(x\) is a fixed, real number larger than -1, and then calling the given formula \(P(n)\), and evaluating \(P(n)\) at the base case \(n=2\).

This gives \((1+x)^2\geq{1+2x}\) which can be rewritten as \(1+2x+x^2\geq{1+2x}\).

It is know that for all real \(x\), the statement \(x^2\geq{0}\) is true.

Here is where I get tripped up.

We need to assume that \(m=n\) a.k.a. \(P(m)\) is true for all natural \(m\geq{2}\).

So we have \((1+x)^m\geq{1+mx}\). Now we need to show that \(P(m+1)\) holds to be true. \(P(m+1)\):

\((1+x)^{m+1}\geq{1+(m+1)x}\).

Now here I would usually try to translate the original formula by plugging in what we had originally, but I am pretty iffy on how to do this with an inequality. If anybody could help me construct a new formula that would help me prove that

\((1+x)^{m+1}\geq{1+(m+1)x}\) is true I would be very thankful.