# inequality proof w/ induction, 2 unknowns

#### skatenerd

##### Active member
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

##### Well-known member
MHB Math Helper
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

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

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

I would consider for the inductive step:

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

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

#### skatenerd

##### Active member
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

Staff member
What I had in mind is to take the equation:

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

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

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

so that we now have:

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

Adding this to $P_m$, we get:

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

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

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

And so we may conclude:

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

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

#### skatenerd

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