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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.

Community
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user2838619
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2 Answers2

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By AM-GM, we have $$(1-x)(1-y)(1-z) \leq \left( \frac{(1-x)+(1-y)+(1-z)}{3} \right)^3 = \frac{8}{27},$$ or $1 - x - y - z + xy + yz + zx - xyz \leq \frac{8}{27}$, hence $xyz + \frac{8}{27} \geq xy+yz+zx$. Note that applying AM-GM is allowed as the numbers $1-x$, $1-y$ and $1-z$ are all positive.

user133281
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Hint: $$(1-x)(1-y)(1-z) = (x+y)(y+z)(z+x)=xy+yz+zx-xyz.$$

Quang Hoang
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