1. ## cauchy-schwarz problem

by considering suitable vectors in R4, show that only one choice of real numbers x,y,z satisfies :

3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0

and find these numbers

2. Define: $
\vec v = \left( {x,y,z,2} \right)
$

Thus the equation turns into: $
3 \cdot \left( {\vec v} \right)^2 = 2 \cdot \left[ {xy + xz + yz + 2x + 2y + 2z} \right]
$

Note that: $
xy + xz + yz + 2x + 2y + 2z = \frac{{\left( {x + y + z + 2} \right)^2 - \left( {\vec v} \right)^2 }}
{2}

$

Thus we have : $
3 \cdot \left( {\vec v} \right)^2 = \left( {x + y + z + 2} \right)^2 - \left( {\vec v} \right)^2
$
or: $
4 \cdot \left( {\vec v} \right)^2 = \left( {x + y + z + 2} \right)^2
$

Consider: $
\vec w = \left( {1,1,1,1} \right)
$

Our equality is: $
\left( {\vec w} \right)^2 \cdot \left( {\vec v} \right)^2 = \left( {\vec w \cdot \vec v} \right)^2

$

By Cauchy-Schwarz we have: $
\left( {\vec w} \right)^2 \cdot \left( {\vec v} \right)^2 \geqslant \left( {\vec w \cdot \vec v} \right)^2
$
with equality iff the vectors are colinear

So this must be the case and therefore $x=y=z=2$ is the only solution to this equation

3. Originally Posted by szpengchao
by considering suitable vectors in R4, show that only one choice of real numbers x,y,z satisfies :

3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0

and find these numbers
If it were not for the vectors idea, we could have simply observed:
$3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0$
$\Leftrightarrow (x-2)^2 + (y-2)^2 + (z-2)^2 + (x-y)^2 + (y-z)^2 + (z-x)^2 = 0$

But PaulRS....awesome