**Problem ** 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

## 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**Book Suggestions:**Secrets in Inequalities

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