Cách 1. Áp dụng bđt Bunhiacopxki : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(\sqrt{a.\frac{1}{a}}+\sqrt{b.\frac{1}{b}}+\sqrt{c.\frac{1}{c}}\right)^2=\left(1+1+1\right)^2=9\)
Cách 2. Áp dụng bđt Cauchy :
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)=> dpcm
A=(a+b+c)(1/a+1/b+1/c)\(\ge\) 9
=a/a + a/b + a/c + b/a + b/b + b/c + c/a + c/b + c/c -9 \(\ge\) 0
=3+ (a/b+b/a) + (a/c+c/a) + (b/c+c/b) - 9 \(\ge\) 0
Ap dung a/b + b/a \(\ge\) 2
A=3+2+2+2+2-9 \(\ge\) 0 (luon dung)