Question

Cho a,b,c>0 và a+b+c=1. Tìm GTLN của\(A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}\)

Answer 1

\(A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}\)

\(=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{\left(1+1+2\right)^2}{a+b+c}=3-16=-13\)có GTNN là - 13

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{4};c=\frac{1}{2}\)

Answer 2

A=\frac{a-1}{a}+\frac{b-1}{b}+\frac{c-4}{c}=1-\frac{1}{a}+1-\frac{1}{b}+1-\frac{4}{c}A=aa−1​+bb−1​+cc−4​=1−a1​+1−b1​+1−c4​

=3-\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\le3-\frac{\left(1+1+2\right)^2}{a+b+c}=3-16=-13=3−(a1​+b1​+c4​)≤3−a+b+c(1+1+2)2​=3−16=−13có GTNN là - 13

Dấu "=" xảy ra \Leftrightarrow a=b=\frac{1}{4};c=\frac{1}{2}⇔a=b=41​;c=21​