Ta có: \(A=\left(x+y\right).1=\left(x+y\right).\left(\frac{2017}{x}+\frac{2018}{y}\right)=2017+2018.\frac{x}{y}+2017.\frac{y}{x}+2018\)
\(\Leftrightarrow A=4035+2017\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{x}{y}\ge4035+2017.2+\frac{x}{y}\)
\(\Leftrightarrow A\ge8069+\frac{x}{y}\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{x}{y}=\frac{y}{x}\Leftrightarrow x^2=y^2\Leftrightarrow x=y=4035\)( thỏa đề bài )
\(\Leftrightarrow minA=8069+1=8070\)
có cả làm bất đẳng thức kiểu này nữa à :)))
\(A=1\left(x+y\right)=\left(\frac{2017}{x}+\frac{2018}{y}\right)\left(x+y\right)\)
\(\ge\left(\sqrt{\frac{2017}{x}.x}+\sqrt{\frac{2018}{y}.y}\right)^2=\left(\sqrt{2017}+\sqrt{2018}\right)^2\approx4070\) (BĐT Bunhiacopxki)
Vậy \(A_{min}=4070\Leftrightarrow x=y=4035\)
Nhầm,sửa lại: \(A_{min}=8070\Leftrightarrow x=y=4035\) :v
