Ta có: \(\frac{a}{b}+\frac{b}{a}\le2\)
Dấu bằng xảy ra khi : a=b
=>\(\frac{2021}{2019}+\frac{2019}{2021}< 2\)
có 2=2019/2021+2/2021+1 (1)
2019/2021+2021/2019=2019/2021+2/2019+1 (2)
có 2/2021 < 2/2019 (3)
từ (1)(2)(3) => 2<2019/2021+2021/2019
k đúng cho mk vs
Mk lộn
\(\frac{2019}{2021}+\frac{2021}{2019}>2\)
\(\frac{2019}{2021}+\frac{2021}{2019}=\left(-0\right),00198019851\)
\(\Rightarrow-0,00198019851< 2\)
Mk dùng máy tính nó ghi :
\(\frac{2019}{2021}+\frac{2021}{2019}=2.00000098\)
Nên nó lớn hơn 2
Ok
