The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$ The real numbers $w, x, y, z$ satisfy the equalities $w+x+y+z=0$ and $w^2+x^2+y^2+z^2=1$. Prove that $-1\le wx+xy+yz+zw\le0$
Supposing $wx+xy+yz+zw>0$. I've proven via contradiction that $wx+xy+yz+zw\le0$. 
But suppose $wx+xy+yz+zw<-1$. Then $wx+xy+yz+zw+1<0$. I'm not sure how to factorise the LHS.

We have $a^2+b^2 \geq 2ab$ for any two real numbers $a$ and $b$. So
$$
\begin{aligned}
0=(x+y+z+w)^2 &= x^2+y^2+z^2+w^2+2(xy+yz+zw+wx)+2(xz+wy) \\
&=1+2(xy+yz+zw+wx)+2(xz+wy) \\
&\leq 1+2(xy+yz+zw+wx)+(x^2+z^2+y^2+w^2) \\
&=2+2(xy+yz+zw+wx) 
\end{aligned}
$$
It follows that
$$-1 \leq xy+yz+zx+zw$$
Equality is achieved when $y=w$ and $x=z$. 
This gives two equality cases $\left(\dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{2}, -\dfrac{1}{2}\right)$ and $\left(-\dfrac{1}{2}, \dfrac{1}{2}, -\dfrac{1}{2}, \dfrac{1}{2}\right)$.
