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H

hien_vuthithanh

$P=x^4+2y^4+3z^4=\dfrac{x^4}{1}+\dfrac{y^4}{\dfrac{1}{2}}+ \dfrac{z^4}{\dfrac{1}{3}} \ge \dfrac{(x^2+y^2+z^2)^2}{1+\dfrac{1}{2}+\dfrac{1}{3}}\ge \dfrac{[\dfrac{(x+y+z)^2}{3}]^2}{\dfrac{11}{6}}=\dfrac{54}{11}$
 
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