2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.