How can I prove these two inequalities? 
For the first one I tried the inequality (x1-1)(x2-1)>=0 and I summed it up with the analougus inequalities but I didn't get what I needed. I also tried to prove it's smaller than (n-1)/2 instead of [n/2], but I didn't get it either.
Please help me prove them! Thanks in advance :)
NOTE. [.] is the integer part
 A: As regards the second one, it suffices to show by induction on $n\geq 2$ that for $x_1,\dots,x_n\in [0,1]$,
$$0\leq \prod_{i=1}^n(1-x_i)\leq 1-\sum_{i=1}^n x_i+\sum_{1\leq i<j\leq n}x_ix_j.$$
For the induction step,
$$\begin{align}\prod_{i=1}^{n+1}(1-x_i)&\leq \left( 1-\sum_{i=1}^n x_i+\sum_{1\leq i<j\leq n}x_ix_j\right)(1-x_{n+1})\\
&= 1-\sum_{i=1}^{n+1} x_i+\sum_{1\leq i<j\leq n+1}x_ix_j-\underbrace{x_{n+1}\sum_{1\leq i<j\leq n}x_ix_j}_{\geq 0}\\
&\leq  1-\sum_{i=1}^{n+1} x_i+\sum_{1\leq i<j\leq n+1}x_ix_j.\end{align}$$
A: The hint.
Let $$f(x_1,x_2,...,x_n)=\sum\limits_{k=1}^nx_k-\sum\limits_{k=1}^nx_kx_{k+1},$$
where $x_{n+1}=x_1.$
Thus, for any $i$ $f$ is a linear function of $x_i$.
But the linear function gets a maximal value for extreme value of the variable.
Thus, $$\max_{x_i\in[0,1]}{f}=\max_{x_i\in\{0,1\}}f$$ and the rest is smooth:
$$f=\sum_{k=1}^nx_k(1-x_{k+1}),$$ which says that $f$ does not depend on the substitution $x_k\rightarrow1-x_k.$
Also, $$f\leq\sum_{k=1}^nx_k,$$ where the equality occurs for $$\sum_{k=1}^nx_kx_{k+1}=0.$$
The second inequality we can prove by  induction and by the same idea.
Can you end it now?  
