Toán 9 Min Max

kido2006 said:

1, Đặt [tex]1=x+y+z\geq 3\sqrt[3]{xyz}=3a\Rightarrow 0<a\leq \frac{1}{3}[/tex]
[tex]P=\frac{1}{1-2(xy+yz+zx)}+\frac{2}{xyz}\geq \frac{1}{1-6\sqrt[3]{(xyz)^2}}+\frac{2}{xyz}=\frac{1}{1-6a^2}+\frac{2}{a^3} \geq 57[/tex]
[tex]\Leftrightarrow \frac{-2(3a-1)(57a^4+19a^3-3a^2-3a-1)}{a^3(6a^2-1)}\geq 0[/tex] ( đúng với [tex] 0<a\leq \frac{1}{3}[/tex] )
Dấu = xảy ra khi [tex]x=y=z=\frac{1}{3}[/tex]
2,
Ta có bổ đề sau [tex](x+y+z)(xy+yz+zx)\leq \frac{9}{8}(x+y)(y+z)(z+x)\Rightarrow \frac{9xyz}{(x+y+z)(xy+yz+zx)}\geq \frac{8xyz}{(x+y)(y+z)(z+x)}[/tex]
[tex]\Rightarrow P\geq \frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{8xyz}{(x+y)(y+z)(z+x)}[/tex]
[tex]=\frac{x+y}{y}+\frac{y+z}{z}+\frac{z+x}{x}+\frac{16xyz}{(x+y)(y+z)(z+x)}-3-\frac{8xyz}{(x+y)(y+z)(z+x)}[/tex]
[tex]\geq 4\sqrt[4]{\frac{x+y}{y}.\frac{y+z}{z}.\frac{z+x}{x}.\frac{16xyz}{(x+y)(y+z)(z+x)}}-3-\frac{8xyz}{8xyz}=8-3-1=4[/tex]

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