# Proof by induction: $(1+x)^n > 1 + nx+nx^2$

This is one of the exercises that appears in Apostol's Calculus I. I'm not sure whether what I did is correct.

1. Let $n_1$ be the smallest positive integer $n$ for which the inequality $(1+x)^n > 1 + nx+nx^2$ is true for all $x > 0$. Compute $n_1$ and prove that the inequality is true for al integers $n \geq n_1$.

The first thing I assumed was that the first number for which the inequality holds was $x=1$. Then:

$$2^n > 2n + 1$$

After a little of inspection, one can notice that the inequality is true for all positive integers $n \geq 3$. So $n_1 = 3$.

Proof (by Induction):

$$P(n): (1+x)^n > 1 + nx+nx^2\qquad \text{for all}\ n \geq 3$$

Base Case: $P(3)$

$$(1+x)^3 > 1 + 3x+3x^2$$ $$x^3+3x^2+3x+1 > 3x^2+3x+1$$

which is true.

Inductive Hypothesis: Assume $P(k)$ is true for a positive integer $k\geq 3$:

$$(1+x)^k > 1 + kx + kx^2\qquad (1)$$

Inductive Step: Prove $P(k+1)$:

$$(1+x)^{k+1} > \underbrace{1 + (k+1)x + (k+1)x^2}_\text{a}$$

If we multiply the inequality $(1)$ by $(1+x)$ we get:

$$(1+x)^{k+1} > (1+x)[1 + kx + kx^2]$$

$$(1+x)^{k+1} > \underbrace{kx^3+2kx^2+kx+x+1}_\text{b}$$

Since $a<b$ and $b < (1+x)^{k+1}$, by Transitivity we have that $a < (1+x)^{k+1}$ and hence $P(n)$ is true as asserted.

Is it correct to assume that my first number is $x=1$, although the problem states that it's true for all $x >0$?

• $x$ is a real greater than $0$, so there's no "smallest $x$ for which the inequality holds". Although letting $x=1$ is a good strategy to determine the positive integer $n_1$. – Workaholic Jan 31 '15 at 21:00
• I would start the proof as follows: "The inequality does not hold for $n=2$, since then for $x=1$ it reads $2^2 > 1 + 2 + 2$. We now show that the inequality holds for all $n \geq 3$, thus proving that $n_1=3$..." – user133281 Jan 31 '15 at 21:03
• @Workaholic Yep. $x>0$, but is there no smallest $n$ for which the inequality holds? – Jazz Jan 31 '15 at 21:04
• @Jazz You just proved that such $n_1$ existed and that it was equal to $3$. – Workaholic Jan 31 '15 at 21:06
• @Jazz The error is the phrase, "the first ... was $x=1$." It makes no sense to say $x=1$ is "first". What you can say is, "Assume the inequality holds for $x=1$." No mention about what is the "first" $x$ is needed. – David K Jan 31 '15 at 21:10

Another approach

From the induction hypothesis, we have

$$(1+x)^k > 1+kx+kx^2.$$

In the induction step, notice

$$1+(k+1)x+(k+1)x^2 = 1+kx+kx^2+x(1+x)$$

Now, using the induction hypothesis and add $x(1+x)$ to both sides:

$$(1+x)^k +x(1+x) > 1+kx+kx^2 + x(1+x).$$

Note that $x(1+x)^k > x(1+x)$, and also note that $$(1+x)^{k+1} = (1+x)^k(1+x) = (1+x)^k+x(1+x)^k.$$ Finally,

\begin{align*} (1+x)^{k+1} &= (1+x)^k+x(1+x)^k \\ &> (1+x)^k+x(1+x) \\ &> 1+kx+kx^2+x(1+x) \\ &= 1+(k+1)x+(k+1)x^2. \end{align*}

If we use the binomial expansion of $(1+x)^n$ we get:$$(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+...\tag{1}$$For this expression to be greater then $1+nx+nx^2$ we must satisfy:$$\frac{n(n-1)}{2}\ge n$$From which you should be able to calculate the minimum $n$

• This makes no sense to me. This shows that $\frac{n(n-1)}{2} \geq n$ is a sufficient condition for the inequality to hold, not that it is a necessary condition. In other words, we can only use this to conclude that the statement is true for all $n \geq 3$, not that it is false for $n=1$ and $n=2$. – user133281 Jan 31 '15 at 21:13
• Ah! I see what you mean @user133281 - you are quite right in pointing that out. I guess we could use this result and then reduce $n$ by $1$ to show it is also a necessary condition? – Mufasa Jan 31 '15 at 21:15
• @Mufasa Note that if we use approach Taylor (or binomial approximation by Newton) combined with the principle of mathematical induction we get demonstration in which the principle of mathematical induction is superfluous. – Elias Costa Jan 31 '15 at 21:24
• @Elias - I could not see anything in the "quoted" question (i.e. 7. Let $n_1$ be...) that asked for the use of mathematical induction. – Mufasa Jan 31 '15 at 21:27