# Inequality (RMO 2015 Mumbai Region Solution)

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

## 4 Replies to “Inequality (RMO 2015 Mumbai Region Solution)”

1. ankurkayal says:

please give the solution of a sum from walker and miller

2. ankurkayal says:

PQRS is a square. T is any point on PR. Perpendiculars TA and TB are drawn on PS and PQ respectively.
TW is drawn perpendicular to AQ and RZ is drawn perpendicular to AQ. Prove that AW=ZQ.

3. ankurkayal says:

please give the solution of this sum adopted from walker and miller

4. devang dinesh pagare says:

please solve that second and third question of prmo 2015