Question

tim gtnn

cho a+b+c=6

tim gtnn

A=(a+1)/a+(b+1)/b+(c+4)/c

Answer 1

Lời giải:

Ta có: \(A=\frac{a+1}{a}+\frac{b+1}{b}+\frac{c+4}{c}\)

\(\Leftrightarrow A=1+\frac{1}{a}+1+\frac{1}{b}+1+\frac{4}{c}=3+\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\)

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)(a+b+c)\geq (1+1+2)^2\)

\(\Leftrightarrow \left(\frac{1}{a}+\frac{1}{b}+\frac{4}{c}\right)\geq \frac{4^2}{a+b+c}=\frac{16}{6}=\frac{8}{3}\)

Do đó: \(A\geq 3+\frac{8}{3}=\frac{17}{3}\) hay \(A_{\min}=\frac{17}{3}\)

Dấu bằng xảy ra khi \((a,b,c)=(\frac{3}{2}; \frac{3}{2}; 3)\)