Question

Cho x,y > 0, x + y = 2. Tìm GTNN của P = \(\frac{2020}{x^2+y^2}+\frac{2019}{xy}\)

Answer 1

\(P=\frac{2020}{x^2+y^2}+\frac{2019}{xy}\)

\(P=\frac{2020}{\left(x+y\right)^2-2xy}+\frac{2019}{xy}\)

\(P=\frac{-2020}{2xy-4}+\frac{2019}{xy}\)

\(P=\frac{-1010}{xy-2}+\frac{2019}{xy}\)

Áp dụng bđt AM-GM : \(ab\le\frac{\left(a+b\right)^2}{4}=\frac{4}{4}=1\)

\(P\ge\frac{-1010}{1-2}+\frac{2019}{1}=3029\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=1\)

Answer 2

Bonking cách em nè:)Gọn hơn xíu:v

\(P=\frac{2020}{x^2+y^2}+\frac{1010}{xy}+\frac{1009}{xy}\)\(=2020\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1009}{xy}\)

\(\ge\frac{2020.4}{\left(x+y\right)^2}+\frac{1009}{\frac{\left(x+y\right)^2}{4}}=2020+1009=3029\)

Đẳng thức xảy khi x = y = 1

Vậy..