Question

Cho ba số tự nhiên a,b,c thỏa mãn a+b+c = 18099. Tìm GTNN của \(H=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Answer 1

Lời giải:

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

\(\Leftrightarrow 18099H=\frac{18099}{a}+\frac{18099}{b}+\frac{18099}{c}\)

\(18099H=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}\)

\(18099H=3+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}\)

Áp dụng BĐT Cô- si ta có:

\(\frac{b}{a}+\frac{c}{a}+\frac{a}{b}+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}\geq 6\sqrt[6]{\frac{b}{a}.\frac{c}{a}.\frac{a}{b}.\frac{c}{b}.\frac{b}{c}.\frac{a}{c}}=6\)

\(\Rightarrow 18099H\geq 3+6=9\)

\(\Rightarrow H\geq \frac{9}{18099}=\frac{1}{2011}\) hay \(H_{\min}=\frac{1}{2011}\)

Dấu bằng xảy ra khi \(a=b=c=6033\)