Cho 3 số dương x , y , z sao cho xy + yz + xz =12 Chứng minh răngf x√((( 12+y^2)(12+z^2))/(12+x^2 )) + y√((( 12+x^2)(12+z^2))/(12+y^2 )) + z√((( 12+x^2)(12+y^2))/(12+z^2 )) = 24

$\displaystyle xy+yz+zx=12$
Ta có 
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ \ \ \ 12+x^{2}\\
=x^{2} +xy+yz+zx\\
=x( x+y) +z( x+y)\\
=( x+y)( x+z)
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
12+y^{2}\\
=y^{2} +xy+yz+zx\\
=y( x+y) +z( x+y)\\
=( x+y)( y+z) \ \ 
\end{array}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ \ \ 12+z^{2}\\
\ =z^{2} +xy+yz+xz\\
=z( y+z) +x( y+z)\\
=( y+z)( x+z)
\end{array}$
Khi đó biểu thức 
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\ \ \ \ \ x\sqrt{\frac{\left( 12+y^{2}\right)\left( 12+z^{2}\right)}{12+x^{2}}} +y\sqrt{\frac{\left( 12+x^{2}\right)\left( 12+z^{2}\right)}{12+y^{2}}} +z\sqrt{\frac{\left( 12+x^{2}\right)\left( 12+y^{2}\right)}{12+z^{2}}}\\
=x\sqrt{\frac{( x+y)( y+z) .( y+z)( x+z)}{( x+y)( x+z)}} +y\sqrt{\frac{( x+y)( x+z) .( y+z)( x+z)}{( x+y)( y+z) \ }} +z\sqrt{\frac{( x+y)( x+z) .( x+y)( y+z) \ }{( y+z)( x+z)}}\\
=x.\sqrt{( y+z)^{2}} +y.\sqrt{( x+z)^{2}} +z\sqrt{( x+y)^{2}}\\
=x.( y+z) +y( x+z) +z( x+y) \ \ \ \ \ \ \ \ \ ( x+y >0;x+z >0;y+z >0\ do\ x;y;z >0)\\
=xy+xz+xy+yz+xz+yz\\
=2( xy+yz+xz)\\
=2.12\\
=24
\end{array}$