\(A=\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{x^2+xy+y^2}\)
- Áp dụng bất đẳng thức Caushy, ta có:
\(A\ge\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{x^2+y^2+\dfrac{x^2+y^2}{2}}\)
\(=\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{\dfrac{3}{2}\left(x^2+y^2\right)}\)
\(=\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{2}{3}.\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{x^2}{xy}+\dfrac{y^2}{xy}+\dfrac{2}{3}.\dfrac{xy}{x^2+y^2}\)
- Áp dụng bất đẳng thức Caushy-Schwarz dạng Engel, ta có:
\(A\ge\dfrac{\left(x+y\right)^2}{xy+xy}+\dfrac{2}{3}.\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{x^2+2xy+y^2}{2xy}+\dfrac{2}{3}.\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{x^2+y^2}{2xy}+\dfrac{2}{3}.\dfrac{xy}{x^2+y^2}+1\)
\(=\left[\dfrac{x^2+y^2}{6xy}+\dfrac{2xy}{3\left(x^2+y^2\right)}\right]+\dfrac{x^2+y^2}{3xy}+1\)
- Áp dụng bất đẳng thức Caushy, ta có:
\(A\ge2\sqrt{\dfrac{x^2+y^2}{6xy}.\dfrac{2xy}{3\left(x^2+y^2\right)}}+\dfrac{2xy}{3xy}+1\)
\(=2.\dfrac{1}{3}+\dfrac{2}{3}+1=\dfrac{7}{3}\)
- Dấu "=" xảy ra khi \(x=y\)
- Vậy \(MinA=\dfrac{7}{3}\).
\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{x^2+y^2}=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{3\left(x^2+y^2\right)}{4xy}+\left[\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}\right]\)
\(=\dfrac{3}{4}.\dfrac{2xy}{xy}+2\sqrt{\dfrac{x^2+y^2}{4xy}.\dfrac{xy}{x^2+y^2}}\)
\(=\dfrac{5}{2}\)
