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)
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