Câu 1: Cho 3 số a,b,c >0 thỏa mãn a+b+c=1 .Chứng minh (1+$\frac{1}{a}$)(1+$\frac{1}{b}$)(1+$\frac{1}{c}$) $\geq$ 64 19/11/2021 Bởi Madeline Câu 1: Cho 3 số a,b,c >0 thỏa mãn a+b+c=1 .Chứng minh (1+$\frac{1}{a}$)(1+$\frac{1}{b}$)(1+$\frac{1}{c}$) $\geq$ 64
Áp dụng bất đẳng thức $AM-GM$ ta được: $1 +\dfrac{1}{3a} +\dfrac{1}{3a} +\dfrac{1}{3a}\geq 4\sqrt[4]{\dfrac{1}{27a^3}}$ $1 +\dfrac{1}{3b} +\dfrac{1}{3b} +\dfrac{1}{3b}\geq 4\sqrt[4]{\dfrac{1}{27b^3}}$ $1 +\dfrac{1}{3c} +\dfrac{1}{3c} +\dfrac{1}{3c}\geq 4\sqrt[4]{\dfrac{1}{27c^3}}$ Nhân vế theo vế ta được: $\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right)\geq 64\sqrt[4]{\dfrac{1}{(27abc)^3}}$ Ta lại có: $abc \leq \left(\dfrac{a + b + c}{3}\right)^3 =\dfrac{1}{27}$ $\to 27abc \leq 1$ $\to (27abc)^3 \leq 1$ $\to \dfrac{1}{(27abc)^3}\geq 1$ $\to \sqrt[4]{\dfrac{1}{(27abc)^3}}\geq 1$ $\to 64\sqrt[4]{\dfrac{1}{(27abc)^3}} \geq 64$ $\to \left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right) \geq 64$ Dấu $=$ xảy ra $\Leftrightarrow a = b = c =\dfrac13$ Bình luận
Áp dụng bất đẳng thức $AM-GM$ ta được:
$1 +\dfrac{1}{3a} +\dfrac{1}{3a} +\dfrac{1}{3a}\geq 4\sqrt[4]{\dfrac{1}{27a^3}}$
$1 +\dfrac{1}{3b} +\dfrac{1}{3b} +\dfrac{1}{3b}\geq 4\sqrt[4]{\dfrac{1}{27b^3}}$
$1 +\dfrac{1}{3c} +\dfrac{1}{3c} +\dfrac{1}{3c}\geq 4\sqrt[4]{\dfrac{1}{27c^3}}$
Nhân vế theo vế ta được:
$\left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right)\geq 64\sqrt[4]{\dfrac{1}{(27abc)^3}}$
Ta lại có:
$abc \leq \left(\dfrac{a + b + c}{3}\right)^3 =\dfrac{1}{27}$
$\to 27abc \leq 1$
$\to (27abc)^3 \leq 1$
$\to \dfrac{1}{(27abc)^3}\geq 1$
$\to \sqrt[4]{\dfrac{1}{(27abc)^3}}\geq 1$
$\to 64\sqrt[4]{\dfrac{1}{(27abc)^3}} \geq 64$
$\to \left(1+\dfrac1a\right)\left(1+\dfrac1b\right)\left(1+\dfrac1c\right) \geq 64$
Dấu $=$ xảy ra $\Leftrightarrow a = b = c =\dfrac13$