\(\left(a+b+c\right)\ge3\sqrt[3]{abc}\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{\frac{1}{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}=9\)
Min=9
dấu = xảy ra khi a=b=c=1
Áp dụng bất đẳng thức Cô-si ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\)
=>\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\sqrt[3]{abc\cdot\frac{1}{abc}}\)
\(\ge9\)
=>GTNN=9
Ta có: \(P=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)
Áp dụng BĐT \(\frac{x}{y}+\frac{y}{x}\ge2\forall x,y\ne0\)
\(\Rightarrow P\ge3+2+2+2=9\)
Dấu "=" xảy ra khi: \(a=b=c\)
