Question: Reading a book I saw this inequality 
$$
(1-x)^n \leq 1 -xn+\frac{n(n-1)}{2}x^2
$$
when $0\leq x\leq 1$ and following the author it descended from the inclusion-exclusion principle. I don't understand why. Moreover is there a simple proof of this inequality?   

Thanks

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I've tried to prove the inequality in the following way:   

From the binomial theorem:
$$
(1-x)^n = \sum\_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^k=1 -xn+\frac{n(n-1)}{2}x^2+\sum\_{k=3}^{n}{{n}\choose{k}}(-1)^{k}x^k
$$
and so we need to prove that:
$$
\sum\_{k=3}^{n}{{n}\choose{k}}(-1)^{k}x^k\leq0
$$
when $0\leq x\leq1$.   

I've tried to use the inequality 
$$
\sum\_{k=3}^{n}{{n}\choose{k}}(-1)^{k}\leq0
$$
that I can prove from $0=(1-1)^n$ and then I've tried to group some terms in the sum but without success.

Answer: 
If true for some $n$, then
\begin{align}
(1-x)^{n+1}&\le(1-x)\left(1-nx+\frac{n(n-1)}2x^2\right)\\
&=1-(n+1)x+\frac{(n+1)n}2x^2-\frac{n(n-1)}2x^3\\
&\le1-(n+1)x+\frac{(n+1)n}2x^2.
\end{align}
and now use induction.
For $n=2$ the proof is trivial, we consider $n>2$.

Let $f(x) = (1-x)^n$ and $g(x)=1-nx+\frac{n(n-1)}{2}x^2$. You have clearly 
$$f(0) = g(0),\quad f'(0) = g'(0),\quad f''(0)=g''(0)$$

However, the third derivative writes
$$ f'''(x) = -n(n-1)(n-2) (1-x)^{n-3} < 0 = g'''(x).$$
Given that $f''(0)=g''(0)$ it is easy to see that $\forall x\in[0,1]\, f'(x)\le g'(x)$. Now use $f(0)=g(0)$ to obtain that $\forall x\in[0,1]\, f(x)\le g(x)$.