Question

cho a,b,c thỏa mãn : a+b+c =1 

Chứng minh : \(\left(1+\frac{1}{a}\right)\times\left(1+\frac{1}{b}\right)\times\left(1+\frac{1}{c}\right)\ge64\)

Answer 1

Cách khác: Áp dụng BĐT AM-GM ta có: 

\(1+\frac{1}{a}=\frac{1}{a}\left(a+b+c+a\right)\ge\frac{1}{4}4\sqrt[4]{a^2bc}\)

\(\Rightarrow1+\frac{1}{a}\ge\frac{4}{a}\sqrt[4]{\frac{a^4bc}{a^2}}=4\sqrt[4]{\frac{bc}{a^2}}\)

Tương tự cũng có: \(1+\frac{1}{b}\ge4\sqrt[4]{\frac{ca}{b^2}};1+\frac{1}{c}\ge4\sqrt[4]{\frac{ab}{c^2}}\)

\(\Rightarrow VT\ge4\sqrt[4]{\frac{bc}{a^2}}4\sqrt[4]{\frac{ca}{b^2}}4\sqrt[4]{\frac{ab}{c^2}}=64\)

Còn tỷ tỷ cách đây cần thì IB nhé !!

Answer 2

Ta cần chứng minh \(\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

\(\Leftrightarrow1+abc+ab+bc+ca+a+b+c\ge1+3\sqrt[3]{\left(abc\right)^2}+3\sqrt[3]{abc}+abc\)

\(\Leftrightarrow ab+bc+ca+a+b+c\ge3\sqrt[3]{\left(abc\right)^2}+3\sqrt[3]{abc}\)

Đúng theo BĐT AM-GM. Thật vậy ta có:

\(\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{abc}\)

\(\ge\frac{\left(1+\sqrt[3]{abc}\right)^3}{abc}\ge64\).Từ \(a+b+c=1\Rightarrow abc\le\frac{1}{27}\)

\(\Rightarrow\frac{\left(1+\sqrt[3]{abc}\right)^3}{abc}=\left(\frac{1}{\sqrt[3]{abc}}+1\right)^3\ge64\)

Đẳng thức xảy ra khi a=b=c=1/3