Proving quadratic inequalities? I am trying to prove that 
$$e^{k+1} ≥ 3 + 3k + k^2$$ with, $$k>2$$
WhatI have done so far:
What we are trying to prove is that $$e^n≥1+n+n^2$$ is a true statement. Since $n=3$ holds, this is our base case. Then, the inductive hypothesis is: $$e^k≥1+k+k^2$$ Hence, I need to show $$e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2$$ is a true statement.

It does not hold.
What about $k=1$? Then $e=e^1 > 1+1+1=3$. It is a contradiction with $e \approx 2.7.$
ADDED:
If $k\ge 3$, then it holds.
Note that $e^x=1+x+\frac {x^2}{2}+\frac {x^3}{3!}+ \cdots+\frac {x^n}{n!}+\cdots$.
So $e^k-(1+k+k^2)>\frac{k^3}{3!}-\frac{k^2}{2}=k^2(\frac k6-\frac12)\ge 0$
