I am trying to verify that $xyz -xy-xz-zy \geq -\frac {8}{27}$ when $x+y+z=1$ and $x,y,z$ positive, preferably without resolving to calculus.
It seems like a standard result but I couldn't come up with some basic application of AM-GM or Muirhead's inequality
Any input is welcome
First observe that: $$ (1 - x)(1 - y)(1 - z) = 1 - (x + y + z) + (xy + yz + xz) - xyz = (xy + yz + xz) - xyz $$ Thus, your inequality is equivalent to the following: $$ (1 - x)(1 - y)(1 - z) \leq \frac{8}{27} $$ Indeed this holds, as by AM-GM: $$ (1 - x)(1 - y)(1 - z) \leq \left(\frac{3 - (x + y + z)}{3}\right)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} $$
The inequality is equivalent to $$(xy+yz+zx)(x+y+z) \leq xyz + {8\over27}(x+y+z)^3$$
which simplifies to
$$2xyz+\sum{x^2y}\leq{8\over27}(x+y+z)^3={8\over27}\sum{x^3}+{8\over9}\sum{xy^2}+{16\over9}xyz$$
which simplifies to
$$6xyz+3\sum{xy^2}\leq8\sum{x^3}$$
Note that $$6xyz\leq 2\sum{x^3}$$ $$3\sum{xy^2}\;\leq6\sum{x^3}$$
The inequality is therefore proven.
