Question

cho x,y dương thỏa mãn x+y=\(\dfrac{2022}{2021}\)

Tìm MIN P=\(\dfrac{2021}{x}+\dfrac{1}{2021y}\)

Answer 1

\(P=\dfrac{1}{2021}\left(\dfrac{2021^2}{x}+\dfrac{1}{y}\right)\ge\dfrac{1}{2021}.\dfrac{\left(2021+1\right)^2}{x+y}=\dfrac{1}{2021}.\dfrac{2022^2}{\dfrac{2022}{2021}}=2022\)

\(P_{min}=2022\) khi \(\left(x;y\right)=\left(1;\dfrac{1}{2021}\right)\)