Question

cmr với b>0 ta có 

\(\frac{b}{b^2+1}+\frac{3\left(b^2+1\right)}{2b}\ge\frac{7}{2}\)

Answer 1

\(\frac{b}{b^2+1}+\frac{3\left(b^2+1\right)}{2b}\)

\(=\frac{b}{b^2+1}+\frac{b^2+1}{4b}+\frac{5\left(b^2+1\right)}{4b}\)

\(\ge1+\frac{5\left(b^2+1\right)}{4b}\)

\(=1+\frac{5}{2}+\frac{5\left(b^2-2b+1\right)}{4b}\)

\(=\frac{7}{2}+\frac{5\left(b-1\right)^2}{4b}\ge\frac{7}{2}\)

Dấu = xảy ra khi b = 1

Answer 2

về kiếm bài lớp 8 đi vì ở đây ít lớp 9 lắm bạn ạ