Let $x,y,z \geqslant 0$ such that $x^2+y^2+z^2+xyz=4$, prove that $$\tag{1} xy+yz+zx +\frac{(x-y)^2(y-z)^2(z-x)^2}{8} \leqslant 2 + \frac{x+y+z-1}{2}xyz $$
Here is a brief history about this inequality
The USAMO 2001 stated that
For all $x,y,z \geqslant 0$ such that $x^2+y^2+z^2+xyz=4$ then
$$\tag{2}xy+yz+zx-xyz \leqslant 2$$
There are several solutions for this inequality. The natural approach is to use the trig substitution $x=2cos (A), y= 2cos(B),z=2cos(C)$ and utilized the identity $$\tag{3} cos^2(A)+cos^2(B)+cos^2(C)+2cos(A)cos(B)cos(C)=1$$
About three years ago, an improve version of USAMO 2001 was proposed, due to the guy a.k.a Mudok in here. With the same condition, his problem stated that
$$\tag{4} xy+yz+zx-\frac{x+y+z-1}{2}xyz \leqslant 2$$
Based on the trig sub in $(3)$, one can see that his inequality is indeed stronger than the original.
I think the version I posted is the strongest one. I could not solve it after several attempts.
