\(\dfrac{2}{xy}=\dfrac{4}{2xy}=\dfrac{1}{2xy}+\dfrac{3}{2xy}\)
Ta có: \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2-2xy\ge0\)
\(\Leftrightarrow x^2+y^2-2xy+4xy\ge4xy\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
Hay \(1\ge2xy.2\)
\(\Rightarrow2xy\le\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2xy}\ge\dfrac{1}{\dfrac{1}{2}}=2\)
\(M=\dfrac{2}{xy}+\dfrac{3}{x^2+y^2}=\dfrac{4}{2xy}+\dfrac{3}{x^2+y^2}=\dfrac{1}{2xy}+\dfrac{3}{2xy}+\dfrac{3}{x^2+y^2}\)
\(\ge2+3.\left(\dfrac{1}{2xy}+\dfrac{1}{x^2+y^2}\right)\)
Áp dụng bất đẳng thức Cosy
\(\ge2+3.\left(\dfrac{4}{2xy+x^2+y^2}\right)\)= 2 + 12 = 14
Vậy Min M =14 khi \(x=y=\dfrac{1}{2}\)
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