Question

Cho a,b,c là các số dương có tích bằng 1. Tìm Min của :

\(B=\dfrac{5bc}{a^2b+a^2c}+\dfrac{5ac}{b^2a+b^2c}+\dfrac{5ab}{c^2b+c^2a}\)

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:

\(B=\frac{5bc}{a^2b+a^2c}+\frac{5ac}{b^2a+b^2c}+\frac{5ab}{c^2b+c^2a}\)

\(B=5\left(\frac{\frac{1}{a^2}}{\frac{1}{c}+\frac{1}{b}}+\frac{\frac{1}{b^2}}{\frac{1}{c}+\frac{1}{a}}+\frac{\frac{1}{c^2}}{\frac{1}{a}+\frac{1}{b}}\right)\)\(\geq 5\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{c}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{a}+\frac{1}{b}}\)

hay \(B\geq \frac{5}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Áp dụng BĐT AM-GM:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3\sqrt[3]{\frac{1}{abc}}=3\) do \(abc=1\)

Suy ra \(B\geq \frac{15}{2}\Leftrightarrow B_{\min}=\frac{15}{2}\)

Dấu bằng xảy ra khi \(a=b=c=1\)