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Ta có
$x^2+\frac{1}{8x}+\frac{1}{8x} \ge 3\sqrt[3]{x^2.\frac{1}{8x}.\frac{1}{8x}}=\frac{3}{4}\Rightarrow x^2+\frac{1}{4x} \ge \frac{3}{4} \quad (1)$.
Tuơng tự
$y^2+\frac{1}{4y} \ge \frac{3}{4}\quad (2)$
$z^2+\frac{1}{4z} \ge \frac{3}{4}\quad (3)$
Mặt khác
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{x+y+z} \ge 6 \Rightarrow \frac{3}{4}\left ( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right ) \ge \frac{9}{2}\quad (4)$
Cộng theo từng vế (1),(2),(3),(4) ta được $A \ge \frac{27}{4}$.
Vậy $\min A=\frac{27}{4}\Leftrightarrow x=y=z=\frac{1}{2}$.
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