Ta có :
\(\left\{{}\begin{matrix}x;y>0\\x+y\le1\end{matrix}\right.\)
\(\Leftrightarrow2\sqrt[]{xy}\le x+y\le1\) \(\left(BĐT.Cauchy\right)\)
\(\Leftrightarrow xy\le\dfrac{1}{4}\left(1\right)\)
Theo đề bài :
\(M=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt[]{1+x^2y^2}\)
\(\Leftrightarrow M\ge2\sqrt[]{\dfrac{1}{x}.\dfrac{1}{y}}.\sqrt[]{1+x^2y^2}\)
\(\Leftrightarrow M\ge2\sqrt[]{\dfrac{1}{xy}}.\sqrt[]{1+\left(xy\right)^2}\)
\(\Leftrightarrow M\ge2\sqrt[]{\dfrac{1}{xy}+xy}\)
\(\Leftrightarrow M\ge2\sqrt[]{\dfrac{1}{a}+a}\left(a=xy\right)\left(2\right)\)
Ta lại có :
\(\dfrac{1}{a}+a=\dfrac{15}{16a}+\left(\dfrac{1}{a}+a\right)\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{a}+a\ge\dfrac{15}{16.\dfrac{1}{4}}+2\sqrt[]{\dfrac{1}{16a}.a}\) \(\left(BĐT.Cauchy\right)\)
\(\Leftrightarrow\dfrac{1}{a}+a\ge\dfrac{15}{4}+\dfrac{2}{4}=\dfrac{17}{4}\)
\(\left(2\right)\Leftrightarrow M\ge2.\dfrac{\sqrt[]{17}}{2}=\sqrt[]{17}\)
Dấu "=" xảy ra khi và chỉ khi :
\(xy=\dfrac{1}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy \(GTNN\left(M\right)=\sqrt[]{17}\left(khi.x=y=\dfrac{1}{2}\right)\)
