Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$, a_1+a_2+...+a_n=n$. Is this true? $a_1a_2+a_2a_3+...+a_na_1\leq n$ 
Given $a_1,a_2,...,a_n>0$ where $n\in\mathbb N$$,  a_1+a_2+...+a_n=n$. Is this true?
$$a_1a_2+a_2a_3+...+a_na_1\leq n$$

By observing:
When $n=1$, this is trivial;
When $n=2$, $ab\leq(\frac {a+b} 2)^2=1\leq2$;
When $n=3$, $ab+bc+ca\leq(\frac {a+b} 2)^2+(\frac {b+c} 2)^2+(\frac {c+a} 2)^2=\frac 1 2(a+b+c)^2-\frac 1 2(ab+bc+ca)$
$\Rightarrow ab+bc+ca\leq3$;
When $n=4$, $ab+bc+cd+da=(a+c)(b+d)\leq(\frac{a+b+c+d} 2)^2=4$.
But I can't find a more general way to prove these at once. Please help.
Thank you.

The inequality does not hold in general. Indeed, take $a_1=a_2=\frac{n}{2}-1$ and 
$a_3=..=a_n=\frac{2}{n-2}.$ Then the left hand side of our inequality is greater that $n^2/4-n+1$ which can be made greater than $n.$
