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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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