Question

cho x,y>0.tim gtnn cua D=x/y +y/x +xy/(x^2+xy+y^2)

Answer 1

Lời giải:
\(D=\frac{x}{y}+\frac{y}{x}+\frac{xy}{x^2+xy+y^2}=\frac{x^2+y^2}{xy}+\frac{xy}{x^2+xy+y^2}\)

\(=\frac{x^2+xy+y^2}{xy}+\frac{xy}{x^2+xy+y^2}-1\)

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

Áp dụng BĐT Cô-si:

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

\(x^2+y^2\geq 2xy\Rightarrow \frac{8(x^2+xy+y^2)}{9xy}\geq \frac{8.3xy}{9xy}=\frac{8}{3}\)

\(\Rightarrow D\geq \frac{2}{3}+\frac{8}{3}-1=\frac{7}{3}=D_{\min}\)

Dấu "=" xảy ra khi $x=y$