<s>[INST] 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.
 [/INST] 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 $k\ge1$
</s>