Áp dụng BĐT Cauchy cho các số dương , ta có :
\(a+\dfrac{1}{4a}\text{ ≥}2\sqrt{a.\dfrac{1}{4a}}=2.\dfrac{1}{2}=1\)
\(b+\dfrac{1}{4b}\text{ ≥}2\sqrt{b.\dfrac{1}{4b}}=2.\dfrac{1}{2}=1\)
\(c+\dfrac{1}{4c}\text{ ≥}2\sqrt{c.\dfrac{1}{4c}}=2.\dfrac{1}{2}=1\)
⇒ \(a+b+c+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥}3\)
⇔ \(a+b+c+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{ ≥}3+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{ ≥ }3+\dfrac{3}{4}.\dfrac{\left(1+1+1\right)^2}{a+b+c}=3+\dfrac{3}{4}.\dfrac{9}{a+b+c}\text{ ≥}3+\dfrac{3}{4}.\dfrac{9}{\dfrac{3}{2}}=\dfrac{15}{2}\) ⇒ \(A_{MIN}=\dfrac{15}{2}."="\text{⇔}a=b=c=\dfrac{1}{2}\)
