By AM-HM, $\sum \frac{1}{x^2} \geq 4$. Hence due to the second equation, $4 \geq 5 - \frac{1}{(xyzw)^2}$ and by AM-GM $1 \geq \sqrt{xyzw} \implies 1 \geq (xyzw)^2$. So $4 \geq 5 - \frac{1}{(xyzw)^2} \geq 4 \implies (xyzw)^2 = 1$. Equality in AM-GM is only possible if $x^2 = y^2 = z^2 = w^2 = 1$ so $x, y, z, w \in \{ 1, -1 \}$ are the only 4-tuples.

step 1 is wrong

Lukas8r20 wrote: step 1 is wrong The conclusion is still true. Let $u = (x, y, z, w), v = (1/x, 1/y, 1/z, 1/w).$ By Cauchy-Schwarz, $\sum\limits_{\text{cyc}} \frac{1}{x^2} = ||v||^2 \ge \frac{|u \cdot v|^2}{||u||^2} = \frac{16}{4} = 4.$

Ah oops i am stupid