[Toán 9] Tìm cực trị

V

viethoang1999



$\dfrac{a^3}{(1+b)(1+c)}+\dfrac{1+b}{8}+\dfrac{1+c}{8}\ge \dfrac{3}{4}a$

Tương tự:
\Rightarrow $P\ge \dfrac{3}{4}\sum a\ge \dfrac{3}{4}.3\sqrt[3]{abc}=\dfrac{9}{4}$
 
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