If $x,y,z>0$ and $x+y+z=1$ then show that $\displaystyle2<(1+x)(1+y)(1+z)<\frac{64}{27}$.

I have solved the right hand first using AM-GM inequality,

$\displaystyle\frac{1+x+1+y+1+z}{3} > \sqrt[3]{(1+x)(1+y)(1+z)}$

$\displaystyle \frac{64}{27}>(1+x)(1+y)(1+z)$

But now I am stuck in proving the left hand side.How do I prove the left-hand side inequality?

  • $\begingroup$ I have seen it somewhere... May be in some Olympiad question. @FBR, Where did you find it? $\endgroup$ – pushpen.paul Oct 17 '14 at 18:30
  • $\begingroup$ Maybe, I don't know.I found this in one of the local textbook. $\endgroup$ – FBR Oct 17 '14 at 18:32

Expand it out. Four of the terms sum to 2, and the remaining terms are positive.

$$(1+x)(1+y)(1+z) = 1 + x + y + z + xy + xz + yx + xyz > 1 + x + y + z = 2$$


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