Toán 9 Min Max

Lê.T.Hà said:

[tex](x-y)^2(x^2+xy+y^2)\geq 0\Leftrightarrow x^4+y^4\geq xy(x^2+y^2)[/tex]
[tex]\Rightarrow P= \sum \dfrac{1}{x^4+y^4+z}\leq \sum \dfrac{1}{xy(x^2+y^2)+z}=\sum\dfrac{z}{xyz(x^2+y^2)+z^2}=\dfrac{x+y+z}{x^2+y^2+z^2}[/tex]
[tex]P \leq \dfrac{3(x+y+z)}{(x+y+z)^2}=\dfrac{3}{x+y+z}\leq \dfrac{3}{3\sqrt[3]{xyz}}=1[/tex]

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