Áp dụng BĐT Cauchy Schwars:

$(x.1+y.1+z.1)^2\le (x^2+y^2+z^2)(1^2+1^2+1^2)\\\to x+y+z\le \sqrt{3(x^2+y^2+z^2)}\\\to S\le \dfrac{xyz.\sqrt{x^2+y^2+z^2}(\sqrt{3}+1)}{(\sqrt{x^2+y^2+z^2})^2(xy+yz+xz)}=\dfrac{xyz(\sqrt{3}+1)}{\sqrt{x^2+y^2+z^2} (xy+yz+xz)}$

Áp dụng BĐT Cauchy:

$\sqrt{x^2+y^2+z^2}\ge \sqrt{3\sqrt[3]{(xyz)^2}}, xy+yz+xz\ge 3\sqrt[3]{(xyz)^2}\\\to \sqrt{x^2+y^2+z^2}(xy+yz+xz)\ge 3\sqrt{3} xyz\\\to S\le \dfrac{\sqrt{3}+1}{3\sqrt{3}}$

Dấu "=" xảy ra khi: $x=y=z$

`S = `$\text{$\dfrac{xyz(x + y + z + \sqrt{x^{2} + y^{2} + z^{2}} )}{(x^{2} + y^{2} + z^{2} )(xy + yz + xz )}$}$

`=> S^2 =` $\text{$\dfrac{x^{2}y^{2}z^{2}(x + y + z + \sqrt{x^{2} + y^{2} + z^{2}} )^{2}}{(x^{2} + y^{2} + z^{2} )^{2}(xy + yz + xz )^{2}}$}$

$\text{Xét mẫu : Áp dụng bất đẳng thức Cô-si ta có :}$

`( x^2 + y^2 + z^2 ) . ( xy + yz + zx ) . ( xy + yz + zx )` `\ge` `27 .`$\text{$\sqrt[3]{(x^{2} . y^{2} . z^{2} )^3}$}$ `= 27 . x^2 . y^2 . z^2 `

`=> S^2 . 27 =` $\text{$\dfrac{(x + y + z + \sqrt{x^{2} + y^{2} + z^{2}} )^{2}}{x^{2} + y^{2} + z^{2} }$}$

`= ((x + y + z )^2)/(x^2 + y^2 + z^2 ) + 1 + ` $\text{$\dfrac{2(x + y + z ) . \sqrt{x^{2} + y^{2} + z^{2}}}{x^{2} + y^{2} + z^{2} }$}$

Ta có `:` `3(x^2 + y^2 + z^2 ) \ge (x+y+z)^2`

`=> 27 . S^2 \le 3 + 1 + 2sqrt{3} = 4 + 2sqrt{3}`

`=> S^2 \le ( 4 + 2sqrt{3} )/27`

`=>` Dấu "=" xảy ra `<=> x = y = z `