Prove that $(1-x)^n \leq 1 -xn+\frac{n(n-1)}{2}x^2$ when $0\leq x\leq 1$ and $n\geq2$ 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


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. 
 A: 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.
A: 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 $(1-x)^n \leqslant 1-nx+\frac12n(n-1)x^2$ 
A: 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 $\forall x\in[0,1]\, f(x)\le g(x)$.
A: 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.
