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Math Help - cauchy-schwarz problem

  1. #1
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    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
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  2. #2
    Super Member PaulRS's Avatar
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    Define: <br />
\vec v = \left( {x,y,z,2} \right)<br />

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

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

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

    Consider: <br />
\vec w = \left( {1,1,1,1} \right)<br />

    Our equality is: <br />
\left( {\vec w} \right)^2  \cdot \left( {\vec v} \right)^2  = \left( {\vec w \cdot \vec v} \right)^2 <br /> <br />

    By Cauchy-Schwarz we have: <br />
\left( {\vec w} \right)^2  \cdot \left( {\vec v} \right)^2  \geqslant \left( {\vec w \cdot \vec v} \right)^2 <br />
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
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  3. #3
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    Quote Originally Posted by szpengchao View Post
    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
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