Toán 9 Tìm giá trị lớn nhất

Erwin Schrödinger said:

[tex]x^3+y^3+z^3=3\geq \frac{(x+y+z)^3}{9}<=>3\geq x+y+z[/tex]
[tex]x^3+y^3+z^3-3xyz=3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)[/tex]
[tex]3A=9(xy+yz+xz)+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)-3\leq 9(xy+yz+xz)+3(x^2+y^2+z^2)-3(xy+yz+xz)-3=6(xy+yz+xz)+3(x^2+y^2+z^2)-3=3(x+y+z)^2-3=3.3^2-3=24[/tex]
=> [tex]3A \leq 24 => A \leq 8[/tex]

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Erwin Schrödinger said:

[tex]x^3+y^3+z^3=3\geq \frac{(x+y+z)^3}{9}<=>3\geq x+y+z[/tex]
[tex]x^3+y^3+z^3-3xyz=3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)[/tex]
[tex]3A=9(xy+yz+xz)+(x+y+z)(x^2+y^2+z^2-xy-yz-xz)-3\leq 9(xy+yz+xz)+3(x^2+y^2+z^2)-3(xy+yz+xz)-3=6(xy+yz+xz)+3(x^2+y^2+z^2)-3=3(x+y+z)^2-3=3.3^2-3=24[/tex]
=> [tex]3A \leq 24 => A \leq 8[/tex]

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