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

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

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

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

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

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

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

So this must be the case and therefore $\displaystyle x=y=z=2$ 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:
$\displaystyle 3(x^2+y^2+z^2+4) - 2( yz+zx+xy) -4(x+y+z)=0$
$\displaystyle \Leftrightarrow (x-2)^2 + (y-2)^2 + (z-2)^2 + (x-y)^2 + (y-z)^2 + (z-x)^2 = 0$

But PaulRS....awesome