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truongduong9083

$P = \dfrac{x^4}{x^2+2xy+3xz}+\dfrac{y^4}{y^2+2yz+3yz}+\dfrac{z^4}{z^2+2zx+3zy} \geq \dfrac{(x^2+y^2+z^2)^2}{x^2+y^2+z^2+5(xy+yz+zx)} \geq \dfrac{1}{1+5(x^2+y^2+z^2)} = \dfrac{1}{6}$
Dấu "= " xảy ra khi $x = y = z = \dfrac{1}{\sqrt{3}}$
 
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