Question

Cho \(P=a+\dfrac{b}{\left(a-b\right)\left(b+1\right)^2}\) với a>b>0.Tìm MinP

Answer 1

T chứng minh với tử bằng 4 :v, còn bằng b thì thua

\(P=a+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\)

\(2P=2a+\dfrac{8}{\left(a-b\right)\left(b+1\right)^2}=2a+\dfrac{16}{2\left(a-b\right)\left(b+1\right)^2}\)

\(=2\left(a-b\right)+b+1+b+1+\dfrac{16}{2\left(a-b\right)\left(b+1\right)^2}-2\ge4\sqrt[4]{2\left(a-b\right).\left(b+1\right).\left(b+1\right).\dfrac{16}{2\left(a-b\right)\left(b+1\right)^2}}-2=6\)

\(\Rightarrow P\ge3\)

Answer 2

Trên tử là số 4 hay b vậy :v