<s>[INST] Prove or disprove $xyz+\frac{8}{27}\ge xy+yz+zx$ if $x+y+z=1$ 
if $x,y,z$ are positive and $x+y+z = 1$,Prove:$$xyz+\frac{8}{27}\ge xy+yz+zx$$
Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: I could rewrite question as:$$(x^2+y^2+z^2)+2xyz+\frac{16}{27}\ge 2(xy+yz+zx)+(x^2+y^2+z^2)=(x+y+z)^2=1$$
it remains to prove $(x^2+y^2+z^2)+2xyz\ge\frac{11}{27}$.by the well-known inequality that $3(x^2+y^2+z^2)\ge(x+y+z)^2=1$ So $x^2+y^2+z^2\ge\frac{1}{3}=\frac{9}{27}$.So it remains to Prove $xyz \ge \frac{1}{27}$.but obviously If we take $x,y$ small enough it turns to be wrong.So there are two possibilities. The question is wrong or I made a mistake on previous steps.
 [/INST] Hint: 
$$(1-x)(1-y)(1-z) = (x+y)(y+z)(z+x)=xy+yz+zx-xyz.$$
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