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: 
Interpretation in terms of inclusion-exclusion principle is not too difficult either, though it could become verbose. Consider a coin with probability $x \in (0, 1)$ of turning Heads. Suppose you want to find the probability of getting only Tails when tossing $n$ such coins. The probability is obviously $(1-x)^n$ by independence of the events.

OTOH, we can look at the probability of all events, i.e. $1$, and reduce from it the probability of all events where there is a Head in each coin individually. This leads to $1-nx$, as there are $n$ coins. However the second term clearly double counts cases where two Heads simultaneously appears, so we add back these, i.e. $\binom{n}{2}x^2$, to get $1-nx+\frac12n(n-1)x^2$. We notice that the last term has double counted all cases where three coins had Heads simultaneously, so this is an overestimate, so $(1-x)^n \leqslant 1-nx+\frac12n(n-1)x^2$
You can apply Taylor around 0. Let $f = (1-x)^n$, then $f'(0) = -n$, $f''(0) = n (n-1)$, and $f'''(\xi) = -n(n-1)(n-2)(1 - \xi)^{n-3}$.

$$(1-x)^n = 1 - nx + \frac{n(n-1)}{2} x^2 + \frac{f'''(\xi)}{6} \, x^3,$$

for some $\xi \in [0, x]$. Since $x \leq 1$, $f'''(\xi) \leq 0$, and so the remainder term is non-positive.