<s>[INST] 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.


*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 $x >0$?
 [/INST] 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,
$$\begin{align*}
(1+x)^{k+1} &= (1+x)^k+x(1+x)^k \\
 &> (1+x)^k+x(1+x) \\
 &> 1+kx+kx^2+x(1+x) \\
 &= 1+(k+1)x+(k+1)x^2.
\end{align*}$$
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