bất đẳng thức khó wa giúp mình với mình đang cần gấp

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demon311

$a+b+c=1$\Rightarrow $3\sqrt[3]{abc} \le 1$\Rightarrow $abc \le \dfrac{1}{27}$
Ta có: $\dfrac{1}{a}+1=\dfrac{1}{3a}+\dfrac{1}{3a}+\dfrac{1}{3a}+1 \ge 4\sqrt[4]{\dfrac{1}{3^3.a^3}}$
Tương tự:
$\dfrac{1}{b}+1 \ge 4\sqrt[4]{\dfrac{1}{3^3.b^3}}$
$\dfrac{1}{c}+1 \ge 4\sqrt[4]{\dfrac{1}{3^3.c^3}}$
\Rightarrow $(\dfrac{1}{a}+1)(\dfrac{1}{b}+1)(\dfrac{1}{c}+1) \ge 4.4.4\sqrt[4]{\dfrac{1}{3^9.(abc)^3}} \ge 64.\sqrt[4]{\dfrac{1}{27^3.\dfrac{1}{27^3}}}=64$
 
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