a: Theo BĐT Cosi, ta sẽ có:
\(\dfrac{x}{y}+\dfrac{y}{x}>=2\cdot\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)
b: \(M=\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{x^2+y^2}{xy}+\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{3\left(x^2+y^2\right)}{4xy}+\left(\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}\right)\)
Ta có: \(\dfrac{3\left(x^2+y^2\right)}{4xy}>=\dfrac{3}{4}\cdot\dfrac{2xy}{xy}=\dfrac{3}{2}\)
\(\dfrac{x^2+y^2}{4xy}+\dfrac{xy}{x^2+y^2}>=2\cdot\sqrt{\dfrac{x^2+y^2}{4xy}\cdot\dfrac{xy}{x^2+y^2}}=2\cdot\dfrac{1}{2}=1\)
Do đó: \(M>=\dfrac{3}{2}+1=\dfrac{5}{2}\)
Dấu '=' xảy ra khi x=y
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