How to prove $(1+x)^n\geq 1+nx+\frac{n(n-1)}{2}x^2$ for all $x\geq 0$ and $n\geq 1$? I've got most of the inductive work done but I'm stuck near the very end. I'm not so great with using induction when inequalities are involved, so I have no idea how to get what I need...
\begin{align}
(1+x)^{k+1}&\geq (1+x)\left[1+kx+\frac{k(k-1)}{2}x^2\right]\\
&=1+kx+\frac{k(k-1)}{2}x^2+x+kx^2+\frac{k(k-1)}{2}x^3\\
&=1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3
\end{align}
And here's where I have no clue how to continue. I thought of factoring out $kx^2$ from the remaining three terms, but I don't see how that can get me anywhere.

\begin{align*}
&= 1 + (k + 1)x + \frac{k(k+1)}{2}x^2 + \frac{k(k-1)}{2}x^3\\
&\geq 1 + (k + 1)x + \frac{k(k+1)}{2}x^2\\
\end{align*}
That's the inductive step. You're done.
