let $x,y,z,w\neq 0$ are real numbers,and such $$x+y+z+w=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{w}=0$$
show that
$$xyzw>0$$
My idea: let $$x+y+z+w=p=0,xy+xz+xw+yz+yw+zw=q,xyz+xyw+yzw+xzw=r=0,xyzw=s$$ so $$r=xy(z+w)+zw(y+x)=xy(-x-y)+zw(x+y)=(x+y)(zw-xy)=0$$ so $$x+y=0(or) zw=xy$$
Hello,@bof and @ Daniel Fischer , How get a contradiction.?
If $x+y\neq0$, note that $zw=\frac{z+w}{\frac{1}{z}+\frac{1}{w}}=\frac{-(x+y)}{-(\frac{1}{x}+\frac{1}{y})}=xy$.
If $x+y=0$, then $z+w=0$ and $xyzw>0$ obviously.
Assume for a contradiction that $xyzw\lt0$. This means that an odd number, $1$ or $3$, of the variables are negative. Without loss of generality (because we can replace each variable by its negative), we may assume that just one variable, say $w$, is negative. So $x,y,z$ are positive numbers. Note that $$\frac1{x+y+z}=-\frac1w=\frac1x+\frac1y+\frac1z.$$ But $x,y,z$ are positive, so $$\frac1{x+y+z}\lt\frac1x\lt\frac1x+\frac1y+\frac1z.$$ This contradiction completes the proof.
In your terminology with $p=0, q, r=0, s$, we have $x, y, z, w$ as the real roots of the quartic $$t^4+qt^2+s=0$$
Thus we note that if $a$ is a root, so is $-a$. Hence the four roots must be of form $a, -a, b, -b$, and which gives $xyzw > 0$.
We have $y+z=-(w+x)$
and $\dfrac1y+\dfrac1z+\dfrac1w+\dfrac1x=\dfrac{y+z}{yz}+\dfrac{w+x}{wx}=(w+x)\left(\dfrac1{wx}-\dfrac1{yz}\right)$
$=\dfrac{(w+x)(yz-wx)}{wxyz}$
$\implies\dfrac{(w+x)(yz-wx)}{wxyz}=0$
So, if the numerator is zero, $\dfrac1y+\dfrac1z+\dfrac1w+\dfrac1x=0;$
Case $\#1:$
If $w+x=0\iff w=-x,y+z=0\iff y=-z$
$\implies wxyz=(xz)^2>0$
Case $\#2:$
If $yz-wx=0\implies wxyz=(wx)^2>0$
Suppose the equation about $a$ : $(a - \frac{1}{x})(a - \frac{1}{y})(a - \frac{1}{z})(a - \frac{1}{w}) = 0$. If you get rid of the brackets you get something like this : $a^4 + Ma^2 + N = 0$. If you solve it about $a^2$ you should get 2 solutions $r_1, r_2 > 0$ (because the equation about $a$ should have 4 roots). Now observe that $0 < r_1r_2 = \frac{1}{xyzw}$.
