Em lớp 9,m.n nhớ giải dấu "=" nữa nha.
Bài 1: Cho x,y,z,a,b,c>0 và $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$.
Tìm min: $A=xyz,B=x+y+z,C=x^{2}+y^{2}+z^{2}$.
a, $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$
$\Rightarrow 1\geq 3\sqrt[3]{\frac{abc}{xyz}}\Rightarrow xyz\geq 27abc$
Dấu = $\Leftrightarrow \frac{a}{x}=\frac{b}{y}=\frac{c}{z}=\frac{1}{3} \Rightarrow x=3a,y=3b,z=3c$
b, $1=\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\geq \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}}{x+y+z} \Rightarrow x+y+z\geq (\sqrt{a}+\sqrt{b}+\sqrt{c})^{2}$
Dấu = $\Leftrightarrow \frac{\sqrt{a}}{x}=\frac{\sqrt{b}}{y}=\frac{\sqrt{c}}{z}=t, \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$(1)
$\Rightarrow x=t\sqrt{a},y=t\sqrt{b},z=t\sqrt{c}$
Thay vào (1)$\frac{a}{t\sqrt{a}}+\frac{b}{t\sqrt{b}}+\frac{c}{t\sqrt{c}}=1 \Rightarrow t=\sqrt{a}+\sqrt{b}+\sqrt{c}$
$x=\sqrt{a}(\sqrt{a}+\sqrt{b}+\sqrt{c}), y=\sqrt{b}(\sqrt{a}+\sqrt{b}+\sqrt{c}),z=\sqrt{c}(\sqrt{a}+\sqrt{b}+\sqrt{c})$