Ta cs \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(1\ge\frac{9}{a+b+c}\)
\(1\le\frac{a+b+c}{9}\)
\(\frac{a+b+c}{4}\ge\frac{9}{4}\)
'='\(\Leftrightarrow a=b=c=3\)
Khi a=b=c=3
A=\(3\cdot\frac{3^2}{3+3\cdot3}\)=\(\frac{9}{4}\)
A\(\ge\frac{a+b+c}{4}\)
(đpcm)