Question

tìm min \(a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\)biết a+2b+3c>=20

Answer 1

Áp dụng BĐT Cô-si

Ta có \(A=a+b+c+\frac{3}{a}+\frac{9}{2b}+\frac{4}{c}\)

\(=\left(\frac{3a}{4}+\frac{3}{a}\right)+\left(\frac{b}{2}+\frac{9}{2b}\right)+\left(\frac{c}{4}+\frac{4}{c}\right)+\left(\frac{a}{4}+\frac{b}{2}+\frac{3c}{4}\right)\)

\(\Rightarrow A\ge2\sqrt{\frac{3a}{4}.\frac{3}{a}}+2\sqrt{\frac{b}{2}.\frac{9}{2b}}+2\sqrt{\frac{c}{4}.\frac{4}{c}}+\frac{1}{4}\left(a+2b+3c\right)\)

\(\Rightarrow A\ge13\)

Dấu bằng xảy ra khi\(a=2;b=3;c=4\)

Vậy\(MinA=13\Leftrightarrow\left(a;b;c\right)=\left(2;3;4\right)\)