MHB Inequality proof w/ induction, 2 unknowns

[skate_nerd]

I am given a statement to prove: Show (without using the Binomial Theorem) that \((1+x)^n\geq{1+nx}\) for every real number \(x>-1\) and natural numbers \(n\geq{2}\). I am given a hint to fix \(x\) and apply induction on \(n\).
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.

 

[Prove It]

I am given a statement to prove: Show (without using the Binomial Theorem) that \((1+x)^n\geq{1+nx}\) for every real number \(x>-1\) and natural numbers \(n\geq{2}\). I am given a hint to fix \(x\) and apply induction on \(n\).
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.

\displaystyle \begin{align*} \left( 1 + x \right) ^{m + 1} &= \left( 1 + x \right) \left( 1 + x \right) ^m \\ &\geq \left( 1 + x \right) \left( 1 + m\,x \right) \\ &= 1 + \left( m + 1 \right) \, x + m\,x^2 \\ &\geq 1 + \left( m + 1 \right) \, x \end{align*}

 

[MarkFL]

Okay, you have shown the base case is true, and so your induction hypothesis $P_m$ is:

$$(1+x)^m\ge1+mx$$

I would consider for the inductive step:

$$(1+x)^{m+1}-(1+x)^{m}=(1+x)^{m}(1+x-1)=(1+x)^{m}x$$

Can you use the inductive hypothesis to construct from this a weak inequality that you can then add to $P_m$?

 

[skate_nerd]

Thanks to both of you for the responses. So I was actually able to figure out and finish the proof going by what Prove It wrote. But to MarkFL, what do you mean by constructing a "weak inequality"? Does that refer to the point where you find an inequality where you have \(mx^2\geq{0}\)?

 

[MarkFL]

What I had in mind is to take the equation:

$$(1+x)^{m+1}-(1+x)^{m}=(1+x)^{m}x$$

and then use the induction hypothesis (which we have multiplied by $x$) to write:

$$(1+x)^{m}x\ge(1+mx)x$$

so that we now have:

$$(1+x)^{m+1}-(1+x)^{m}\ge(1+mx)x$$

Adding this to $P_m$, we get:

$$(1+x)^{m+1}\ge 1+mx+(1+mx)x=1+(m+1)x+mx^2$$

Since $mx^2\ge0$, we have:

$$(1+x)^{m+1}\ge1+(m+1)x+mx^2\ge1+(m+1)x$$

And so we may conclude:

$$(1+x)^{m+1}\ge1+(m+1)x$$

Thus, we have derived $P_{m+1}$ from $P_{m}$, thereby completing the proof by induction.

 

[skate_nerd]

Ahh I see. Yeah I ended up writing something similar to that, with the same ending. Thanks for the help guys!