Question

Cho a,b,c thoa man a+2b +3c >= 21 cmr 

a+b+c +3/a+9/2b+4/c>=13

Answer 1

mk chỉnh lại đề nhé:  \(a+2b+3c\ge20\)

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

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

\(\ge2\sqrt{\frac{3a}{4}.\frac{3}{a}}+2\sqrt{\frac{b}{2}.\frac{9}{2b}}+2\sqrt{\frac{c}{4}.\frac{4}{c}}+\frac{1}{4}\left(a+2b+3c\right)\) (BĐT AM-GM)

\(\ge\)\(3+3+2+\frac{20}{4}=13\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=2,b=3,c=4\)