Question

Cho x,y>0

Tìm GTNN của:

B= \(\dfrac{x}{y}\)+\(\dfrac{y}{x}\)+\(\dfrac{xy}{x^2+xy+y^2}\)

Answer 1

Lời giải:

Ta có \(B=\frac{x}{y}+\frac{y}{x}+\frac{xy}{x^2+xy+y^2}=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{xy}{x^2+xy+y^2}\)

\(=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}-\frac{1}{9}\)

Áp dụng BĐT AM-GM:

\(\frac{x}{y}+\frac{y}{x}\geq 2\)

\(\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}\geq 2\sqrt{\frac{1}{9}}=\frac{2}{3}\)

Do đó: \(B\geq \frac{8}{9}.2+\frac{2}{3}-\frac{1}{9}=\frac{7}{3}\Leftrightarrow B_{\min}=\frac{7}{3}\)

Dấu bằng xảy ra khi $x=y$