Question

Tìm min:

\(M=4\left(a+b+c\right)^2+3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) \(với\) \(a,b,c>0\)

Answer 1

\(\)\(M=4\left(a+b+c\right)^2+3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge4\left(a+b+c\right)^2+\dfrac{27}{a+b+c}=4\left(a+b+c\right)^2+\dfrac{13,5}{a+b+c}+\dfrac{13,5}{a+b+c}\ge3\sqrt[3]{4.13,5.13,5}=27\Rightarrow min=27\Leftrightarrow a=b=c=\dfrac{1}{2}\)