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.
 A: We need to show $(1+x)^{k+1}\ge1+(k+1)x+\frac{(k+1)k}2x^2$
You already have 
$$
\begin{align}
(1+x)^{k+1}&\geq1+(k+1)x+kx^2+\frac{k(k-1)}{2}x^2+\frac{k(k-1)}{2}x^3
\end{align}$$
$$=1+(k+1)x+x^2\frac{k(k+1)}2+\frac{k(k-1)}{2}x^3$$ 
which is $$\ge1+(k+1)x+x^2\frac{k(k+1)}2$$ if $\displaystyle\frac{k(k-1)}{2}x^3\ge0$ which is true for $x\ge0$ and $k\ge1$
A: If $n=1$, it is trivial. Suppose it is true for $n$. We will show that this formula is true for $n+1$.
$$
\begin{aligned}
(1+x)^{n+1}&=(1+x)^n(1+x)\\
&\ge \left( 1+nx+\frac{n(n-1)}{2} x^2 \right)(1+x)\\
&= 1+nx+\frac{n(n-1)}{2}x^2 +x+nx^2 +\frac{n(n-1)}{2} x^3\\
&= 1+(n+1)x+\frac{n^2-n+2n}{2}x^2+\frac{n(n-1)}{2} x^3\\
&= 1+(n+1)x +\frac{n^2+n}{2}x^2+\frac{n(n-1)}{2} x^3\\
&\ge 1+(n+1)x +\frac{n(n+1)}{2}x^2\\
&= 1+(n+1)x +\frac{(n+1)(n+1-1)}{2}x^2
\end{aligned}
$$
so we get desired result.
A: \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.
