Show that $\displaystyle 2<(1+x)(1+y)(1+z)<\frac{64}{ 27}$

Question

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?

Answer 1

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

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