if x+y+z = 0 then x2y2+y2z2+x2z2/ x4+y4+z4 = ,Answer given was ½ but how it came tell please

(x-y)2>=0; Therefore, x2+y2>=2xt>=0
Also using Nima J's proof, it follows that
(X2-y2)2>=0; Therefore, x4+y4>=2x2y2
Dividing both sides by 2 gives
(a) ½(x2+y2)>=xy and
(b) ½(x4+y4)>=x2y2

Now since (x2+y2)>=2xy; it follows that
(x2+y2)(z2+w2)>=(2xy)(2zw)=4xyzw
Therefore, multiplying the factors gives:
(c) x2z2+x2w2+y2z2+y2w2>=4xyzw

Now
X4+y4+z4+w4 =
½[(x4+z4)+ (x4+w4)+ (y4+z4)+ (y4+w4)]>=
[(x2z2)+ (x2w2)+ (y2z2)+ (y2w2)]>= 4xyzw; equations (b) &(c)
Therefore,
X4+y4+z4+w4>=4xyzw