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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 . Prove that (1+x)(1+y)(1+z) \le 4 + 4xyz

Discussion

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

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

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

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

PROOF 2 (suggested by Arkabrata Das)

CHECK FILE: new doc 1720151229235503023

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