I'm stuck on this proof. I assume it's supposed to be an induction proof, but I cannot figure out how to algebraically prove that $(1+x)^{k+1}\geq \frac{(k+1)(k)(k-1)}{6}x^3$. Any help would be appreciated.
Note: it's easy to prove that $(1+x)^3\geq \frac{3(3-1)(3-2)}{6}x^3$, since the right hand side is equal to $x^3$, and it's easy to see that $(1+x)^3\geq x^3$.
I'm stuck on understanding how to relate the assumption that $(1+x)^k\geq \frac{k(k-1)(k-2)}{6}x^3$ to the inequality needed, that $(1+x)^{k+1}\geq \frac{(k+1)k(k-1)}{6}x^3$, in order to prove that $P(k)\Rightarrow P(k+1)$ is true.
Hint If $x$ is positive, this follows immediatelly from the Binomial theorem.
If you want to use induction, you can prove easily the stronger statement $$(1+x)^k \geq 1+kx+\frac{k(k-1)}{2}x^2+\frac{k(k-1)(k-2)}{6}x^3 \hskip{1cm} \forall k \geq 3$$
Added: The inductive step
$$(1+x)^{k+1}=(1+x)^k(1+x) \geq (1+kx+\frac{k(k-1)}{2}x^2+\frac{k(k-1)(k-2)}{6}x^3)(1+x)$$ with the inequality following from $P(k)$ and $1+x \geq 0$.
Now, just expand the RHS, and use the fact that the term $\alpha x^4 \geq 0$.