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

Question

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

7. 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 $n \geq n_1$.

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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 $x >0$?

Answer 1

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 $n$

Answer 2

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,

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