11This was more confusing than anything - it always diverted me towards wrong paths!
Symmetry gives us a desired order of x≥y≥z.
So x3y≥z4....(1)
From 1st eqn we have
z4+3≤x3y+3≤4z....
That gives f(z)=z4-4z+3≤0...
But f(z)=(z-1)2(z2+2z+3)≥0, for all real z.
That gives f(z)=0, meaning z=1, since equality holds in (1) for x=y=z, wee have the reqd soln set as (x,y,z)=(1,1,1).
