Giải:
Có:
\(P=\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\)
Vì \(\left(x+\dfrac{1}{x}\right)^2\ge0,\forall x\) và \(\left(y+\dfrac{1}{y}\right)^2\ge0,\forall y\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\ge0;\forall x,y\)
\(\Rightarrow Min_P=0\)
Chúc bạn học tốt!
Áp dụng BĐT \(x^2+y^2\ge\dfrac{1}{2}\left(x+y\right)^2\) và BĐT \(xy\le\dfrac{1}{4}\left(x+y\right)^2\), ta có:
\(\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2\)
\(\ge\dfrac{1}{2}\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)^2\)\(=\dfrac{1}{2}\left(1+\dfrac{x+y}{xy}\right)^2\)
\(\ge\dfrac{1}{2}\left(1+\dfrac{1}{\dfrac{1}{4}\left(x+y\right)^2}\right)^2=\dfrac{25}{2}\left(x+y=1\right)\)
Dấu "=" xảy ra khi x = y = 0,5