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Question

$(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\ge64$
dieu kien $a+b+c=1$

score 4 · Answer 1

Áp dụng BĐT Cauchy ta có:
$1+\frac{1}{a}=\frac{a+1}{a}=\frac{a+a+b+c}{a}\ge\frac{4\sqrt[4]{a^2bc}}{a}$
Tương tự: $1+\frac{1}{b}\ge\frac{4\sqrt[4]{ab^2c}}{b},1+\frac{1}{c}\ge\frac{4\sqrt[4]{abc^2}}{c}$
Suy ra: $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\ge64$
Dấu bằng xảy ra khi: $a=b=c=\frac{1}{3}$.