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[hoanglong9a1]

Cho x, y,z > 0 thỏa mãn xy+yz+zx= $\frac{9}{4}$.Tìm Min A = $x^2+14y^2+10z^2-4\sqrt{2y}$

[trieutuyennham]

Cho x, y,z > 0 thỏa mãn xy+yz+zx= $\frac{9}{4}$.Tìm Min A = $x^2+14y^2+10z^2-4\sqrt{2y}$

 

Ta có

$A=x^{2}+14y^{2}+10z^{2}-4\sqrt{2y}=(\frac{x^{2}}{2}+8y^{2})+\frac{x^{2}}{2}+8z^{2}+2(y^{2}+z^{2})+4y^{2}-4\sqrt{2y}\geq 4(xy+yz+xz)+4y^{2}-4\sqrt{2y}=9+4y^{2}-4\sqrt{2y}=4y^{2}+1-4\sqrt{2y}+8\geq 4y+8-4\sqrt{2y}\geq 2(\sqrt{2y}-1)^{2}+6\geq 6$