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$6=\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z} \ge 6\sqrt[6]{\dfrac{1}{x^3y^2z}}$

\Rightarrow $x^3y^2z \ge 1$

$x^3+y^2+z \ge 3\sqrt[3]{x^3y^2z} \ge 3$

Dấu "=" xảy ra \Leftrightarrow $x=y=z=1$
 
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