Well, the first is pretty straightforward. Note that $x_1x_2=\sqrt{x_1^2x_2^2}\le 1/2(x_1^2+x_2^2)$ and by similar estimates $LHS\le x_1^2+x_2^2+x_3^2$. But adding $2x_1x_2+2x_2x_3+2x_2x_3$ to both sides we get what we needed. EDIT: Another proof: if $x+y+z=0$, we have $0=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$, where $x^2+y^2+z^2\ge0$, so $xy+yz+zx\le0$. Yet another proof: $xy+yz+zx=xy+z(x+y)=xy-(x+y)^2=-(x^2+xy+y^2)=-[(x-y/2)^2+3/4y^2]\le0$.

randomusername wrote: Note that $x_1x_2=\sqrt{x_1^2x_2^2}$ It's wrong! Try $x_1=1$ and $x_2=-1$.

Yes sorry. But the needed ineq still holds: $1/2(x^2+y ^2)\ge xy$ is $1/2(x-y)^2\ge0$.

The second part isn't true: $x_1x_2+x_2x_3+x_3x_4+x_4x_1=(x_1+x_3)(x_2+x_4)=-(x_1+x_3)^2\le0$. In general, we have a problem for even $n$ because by taking $x_i=1$ for odd and $x_i=-1$ for even $i$, the LHS is negative.

There was a typo($\geq 0$) in the second part. I fixed it. Sorry.

Another way to do the second: Let x1 etc the roots of the polynomial p= x^3 + ax+b. Then since x1, x2, x3 are real this polynomial has exactly 3 real roots. then p has 2 turning points, where its derivative 3x^2 + a =0. Then clearly a must be negative or else p would be monotonous and would have only 1 real root.

For the second part, I've found no $n$ obeys the condition. We show that for every $n\geq 4$, there exists an $n-$tuple $(x_1,\dots,x_n)$ for which, even though $x_1+\cdots+x_n=0$, we have, $x_1x_2+\cdots+x_nx_1>0$. First, suppose that $n=2k+1$. Now, consider the following tuple, $(ka,ka,-a,\cdots,-a,-2a,-a,\cdots,-a)$. For this tuple, the sum is indeed $0$, and also, the quantity that we study is strictly positive. Similarly, if $n=2k$, then consider the tuple, $(ka,ka,-a,\cdots,-a,-3a,\cdots,-a)$. One can easily verify that this indeed constitutes as a counterexample.

For $n\geq 5$ it is enough $x_1=x_2=-1,x_4=2$ and $x_i=0$ for $i \neq 1,2,4$