Prove an inequality by induction

Consider the inequality

$(1+x)^n > 1+nx+nx^2 \qquad \text{for } x > 0.$

Find the smallest integer for which this inequality holds, and prove that it holds for all larger integers.

Claim: The inequality

$(1+x)^n > 1+nx+nx^2$

for all $n \geq 3$ (and for no smaller positive integers $n$ ).
Proof. First, we show that it is not true for $n=1$ or $n=2$ . If $n=1$ , then we have

$(1+x) \not> 1+x+x^2$

since $x>0$ implies $x^2 > 0$ , so the right side is larger than the left.
If $n=2$ , then we have

$(1+x)^2 = 1+2x+x^2 \not> 1+2x+2x^2$

since again, $x^2 > 0$ , so $2x^2 > x^2$ . Therefore, the statement is false for both $n=1$ and $n=2$ .
Now, for $n=3$ we have

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

since $x^3 > 0$ . Hence, the statement is true for $n=3$ .
Now assume the statement is true for some $k \in \mathbb{Z}_{\geq 3}$ .` So, we have,

$\begin{align*} &&(1+x)^k &> 1+kx+kx^2 & (\text{Ind. Hyp.})\\ \implies && (1+x)^k (1+x) &> (1+kx+kx^2)(1+x) & (\text{Multiplying by } 1+x > 1)\\ \implies && (1+x)^{k+1} &> 1+kx + kx^2 + x + kx^2 + kx^3 \\ \implies && (1+x)^{k+1} &> 1+(k+1)x +(k+1)x^2. \end{align*}$

Where the final inequality follows since $kx^3 > 0$ and $2kx^2 > (k+1)x^2$ , so we have just made the term on the right smaller.
Thus, if the inequality is true for some $k \geq 3$ , then it is true for $k+1$ . Hence, it is true for all $n \geq 3. \qquad \blacksquare$