How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.

Another possible method is generation of a function and calculating its minimum:
$$f(a,b)=a^4+b^4+2-4 a b$$ 
$$D_a f(a,b)=4 a^3-4 b = 0$$ 
$$ a^3=b$$ 
$$ f(a)=a^4+a^{12}+2-4a^4=a^{12}+2-3a^4$$ 
$$D_a f(a)=12a^{11}-12a^3=0$$ 
$$a^3 (a^8-1)=0$$ 
Minimum must be at one of:
$$f(0,0)=2$$
$$f(1,1)=0$$
$$f(-1,-1)=0$$
so
$$f(a,b)=a^4+b^4+2-4 a b\geq0$$ 
