Even the inequality $(x,z;x,z)^2 \le (x,x;z,z)(z,z;x,x)$ is false:
Let $V = \mathbb{R}^2$, with basis $x,z$. Take $(x,x;x,x) = 100$, $(x,z;x,x)=0$, $(z,z;x,x)=1$, $(x,x;x,z)=0$, $(x,z;x,z)=50$, $(z,z;x,z)=0$, $(x,x;z,z) = 1$, $(x,z;z,z)=0$, $(z,z;z,z)=100$, and extend to all of $V^4$ by symmetry and multilinearity.
To check that positive semi-definiteness holds, note that we just need to check that
$(x+az,x+az;x+bz,x+bz) = 100 + a^2 + 200ab + b^2 + 100a^2b^2 \ge 0$,
which easily follows from AM-GM.
Now note that we have $2500 = (x,z;x,z)^2 > (x,x;z,z)(z,z;x,x) = 1$.
In fact, we even have $6250000 = (x,z;x,z)^4 > (x,x;x,x)(x,x;z,z)(z,z;x,x)(z,z;z,z) = 10000$.
Edit: On the other hand, we can prove the following inequality:
$4(x,y;z,w)^2 \le ((x,x;z,z)+(x,x;w,w))((y,y;z,z)+(y,y;w,w))$.
To see this, note that by positive semi-definiteness we have
$0 \le (x+ay,x+ay;z+w,z+w) + (x-ay,x-ay;z-w,z-w)$ $ = 2((x,x;z,z)+(x,x;w,w)) + 8a(x,y;z,w) + 2a^2((y,y;z,z)+(y,y;w,w))$
for any $a$, and plugging in $a = -\frac{2(x,y;z,w)}{(y,y;z,z)+(y,y;w,w)}$ we get the desired inequality.
