Question

Bài 1: Cho a, b thỏa mãn ab > 2020a + 2021b

Chứng minh rằng: a+b > \(\left(\sqrt{2020}+\sqrt{2021}\right)^2\)

Bài 2: Tìm x,y thỏa mãn \(\sqrt{x-3}+\sqrt{5-x}=y^2+2\sqrt{2019}.y+2021\)

Answer 1

bài 1 ta có 

\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\)  ( BDT Bunhia )

do đó

\(a+b=ab.\left(\frac{1}{a}+\frac{1}{b}\right)\ge\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\)

vậy ta có đpcm.

bài 2.

ta có \(VT=\sqrt{x-3}+\sqrt{5-x}\le2\)( BDT Bunhia )

\(VP=y^2+2.\sqrt{2019}y+2021=\left(y+\sqrt{2019}\right)^2+2\ge2\)

suy ra PT có nghiệm \(\hept{\begin{cases}x-3=5-x\\y+\sqrt{2019}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=-\sqrt{2019}\end{cases}}}\)