\(VT=\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)=\left(1+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{ab}\right)\left(1+\dfrac{1}{c}\right)=1+\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{bc}+\dfrac{1}{a}+\dfrac{1}{ac}+\dfrac{1}{ab}+\dfrac{1}{abc}=1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{abc}\) Áp dụng BĐT Cauchy nhiều lần , ta có :
\(a+b+c\ge3\sqrt[3]{abc}\Leftrightarrow\left(\dfrac{a+b+c}{3}\right)^3\ge abc\Leftrightarrow\dfrac{1}{27}\ge abc\Leftrightarrow\dfrac{1}{abc}\ge27\)
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge3\sqrt[3]{\dfrac{1}{ab}.\dfrac{1}{bc}.\dfrac{1}{ac}}=3\sqrt[3]{\left(\dfrac{1}{abc}\right)^2}\ge3\sqrt[3]{27.27}=27\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\ge3\sqrt[3]{27}=9\)
\(\Rightarrow VT\ge27+9+27+1=64\)
\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
\("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)
Ta có:
\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)=\left(1+\dfrac{a+b+c}{a}\right)\left(1+\dfrac{a+b+c}{b}\right)\left(1+\dfrac{a+b+c}{c}\right)\)
\(=\left(\dfrac{2a+b+c}{a}\right)\left(\dfrac{a+2b+c}{b}\right)\left(\dfrac{a+b+2c}{c}\right)\)
\(=\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)\left(\dfrac{a+b}{b}+\dfrac{b+c}{b}\right)\left(\dfrac{a+c}{c}+\dfrac{b+c}{c}\right)\)
Áp dụng bất đẳng thức Cô - si ta có:
\(\dfrac{a+b}{a}+\dfrac{a+c}{a}\ge2\sqrt{\dfrac{\left(a+b\right)\left(a+c\right)}{a^2}}\)
\(\dfrac{a+b}{b}+\dfrac{b+c}{b}\ge2\sqrt{\dfrac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
\(\dfrac{a+c}{c}+\dfrac{b+c}{c}\ge2\sqrt{\dfrac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\Rightarrow\left(\dfrac{a+b}{a}+\dfrac{a+c}{a}\right)\left(\dfrac{a+b}{b}+\dfrac{b+c}{b}\right)\left(\dfrac{a+c}{c}+\dfrac{b+c}{c}\right)\ge8\sqrt{\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}{a^2b^2c^2}}=8.\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Áp dụng bất đẳng thức Cô - si ta có:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8\sqrt{a^2b^2c^2}=8abc\)
\(8.\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\ge8\sqrt{a^2b^2c^2}=8.\dfrac{8abc}{abc}=64\)
hay \(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\) (đpcm)