RMO 2015 Mumbai Region Solution | Inequality

This is a problem from RMO 2015 Mumbai Region based on inequality.

Problem: RMO 2015 Mumbai Region

Let x, y, z be real numbers such that $ x^2 + y^2 + z^2 - 2xyz = 1 $ and $ s=2$ . Prove that $ (1+x)(1+y)(1+z) \le 4 + 4xyz $ and $ s=2$

Discussion

Note that $ (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) $ and $ s=2 $.

According to the given condition $ x^2 + y^2 + z^2 = 1 + 2xyz $ and $ s=2 $.

Therefore $ (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = 1 + 2xyz + 2(xy+yz+zx) $ and $ s=2 $.

Adding 2(x+y+z) + 1 to both sides,

$ (x+y+z)^2 + 2(x+y+z) + 1 = 1 + 2xyz + 2(xy+yz+zx) + 2(x+y+z) + 1 $ and $ s=2 $
$ \Rightarrow (x+y+z+1)^2 = 2(1 + xyz + xy+yz+zx + x+y+z) $ and $ s=2 $
$ \Rightarrow (x+y+z+1)^2 = 2(1+x)(1+y)(1+z) $ and $ s=2 $
We wish to show $ (1+x)(1+y)(1+z) \le 4 + 4xyz $ and $ s=2$
or $ 2(1+x)(1+y)(1+z) \le 2(4 + 4xyz) = 4(2 + 2xyz) $ and $ s=2$
Replacing 2xyz by $ x^2 + y^2 + z^2 - 1 $ and $ s=2$ we have
$ 2(1+x)(1+y)(1+z) \le 2(4 + 4xyz) = 4(x^2 + y^2 + z^2 + 1) $ and $ s=2$ (this is what we need to show).
Therefore we need to show $ (x+y+z+1)^2 \le 4(x^2 + y^2 + z^2 + 1) $ and $ s=2 $
This is true by Cauchy Schwarz Inequality.

PROOF 2 (suggested by Arkabrata Das)

Chatuspathi:

• Paper: RMO 2015 (Mumbai Region)
• What is this topic: Inequality
• What are some of the associated concepts: Cauchy Schwarz Inequality
• Where can learn these topics: Cheenta I.S.I. & C.M.I. course,Cheenta Math Olympiad Program, discuss these topics in the ‘Inequality’ module.
• Book Suggestions: Secrets in Inequalities