Question

Cho a+2b+3c>=20

Tìm Min A= a+b+c+\(\dfrac{3}{a}\)+\(\dfrac{9}{2b}\)+\(\dfrac{4}{c}\)

Answer 1

\(A=\dfrac{3}{a}+\dfrac{3a}{4}+\dfrac{9}{2b}+\dfrac{b}{2}+\dfrac{4}{c}+\dfrac{c}{4}+\dfrac{a}{4}+\dfrac{b}{2}+\dfrac{3c}{4}\)

Áp dụng bất đẳng thức Cô-si ta được:

\(\dfrac{3}{a}+\dfrac{3a}{4}\ge2\sqrt{\dfrac{3}{a}.\dfrac{3a}{4}}=3\)

\(\dfrac{9}{2b}+\dfrac{b}{2}\ge2\sqrt{\dfrac{9}{2b}.\dfrac{b}{2}}=3\)

\(\dfrac{4}{c}+\dfrac{c}{4}\ge2\sqrt{\dfrac{4}{c}.\dfrac{c}{4}}=2\)

\(\Rightarrow A\ge3+3+2+\dfrac{1}{4}\left(a+2b+3c\right)\)

\(\Rightarrow A\ge8+\dfrac{1}{4}.20=13\)

Vậy Min A=13. Dấu "=" xảy ra \(\Leftrightarrow\) a=2, b=3,c=4