[Toán 9] Tìm cực trị

E

eye_smile

Ta có: $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \ge 9$ (Do $x+y+z=1$)

\Leftrightarrow $xy+yz+xz \ge 9xyz$

\Rightarrow $xy+yz+xz-12xyz \ge -3xyz $

Lại có: $x+y+z \ge 3\sqrt[3]{xyz}$

\Leftrightarrow $xyz \le \dfrac{1}{27}$

\Leftrightarrow $-3xyz \ge \dfrac{-1}{9}$

\Rightarrow $BT \ge \dfrac{-1}{9}$

Dấu "=" xảy ra \Leftrightarrow $x=y=z=\dfrac{1}{3}$
 
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