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[darkevil]

cho $x^{2}+y^{2}+z^{2}+\frac{16}{25}xy=2013$ 

tìm giá trị lớn nhất của P=xy+yz+xz

[Tom Xe Om]

Ta có $\frac{7}{25}(x^2 + y^2)\geq \frac{7}{25}.2xy=\frac{14}{25}xy$

          $\frac{18}{25}x^2 + \frac{z^2}{2}\geq 2\sqrt{\frac{18}{25}x^2\frac{z^2}{2}}=\frac{6}{5}xz$

          $\frac{18}{25}y^2+\frac{z^2}{2}\geq 2\sqrt{\frac{18}{25}y^2\frac{z^2}{2}}=\frac{6}{5}yz$

theo GT: $x^2 + y^2 + z^2 + \frac{16}{25} = \frac{7}{25}(x^2 + y^2)+(\frac{18}{25}x^2 + \frac{z^2}{2})+(\frac{18}{25}y^2 +\frac{z^2}{2})+\frac{16}{25}xy\geq \frac{14}{25}xy+ \frac{6}{5}xz+ \frac{6}{5}yz+ \frac{16}{25}xy=\frac{6}{5}(xy+yz+xz)$

=> $2013\geq \frac{6}{5}(xy+yz+xz)$

=> $xy+yz+xz \leq \frac{3355}{2}$