Đặt \(A=x+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge3\)

\(=\left(x-y\right)+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}+\left(y+1\right)-1\)

Áp dụng BĐT Cô-si cho 2 số dương ta có :

\(\left(x-y\right)+\dfrac{4}{\left(x-y\right)\left(y+1\right)^2}\ge2\sqrt{\dfrac{\left(x-y\right).4}{\left(x-y\right)\left(y+1\right)^2}}=\dfrac{4}{y+1}\)

Xảy ra khi : \(\left(x-y\right)\left(y+1\right)=2\) ( do \(a,b>0\))

\(\Rightarrow A\ge\dfrac{4}{y+1}+\left(y+1\right)-1\)

Sử dụng Cô-Si lần nữa, ta có :

\(\dfrac{4}{y+1}+\left(y+1\right)\ge2\sqrt{\dfrac{4}{y+1}\left(y+1\right)}=2.2=4\)

Xảy ra khi \(\left(y+1\right)^2=4\Leftrightarrow y=1\)

Từ đây ta có thể thấy : \(A\ge4-1=3\)

Dấu "=" xảy ra khi \(\left(x-y\right)\cdot\left(y+1\right)=2\) và \(y=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right..\)

Bài này hồi lúc cũng không biết làm, h biết truyền lại cho bạn :D

1. Ta có: \(a-b+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge\dfrac{4}{b+1}\)

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

lại có: \(\dfrac{4}{b+1}+b+1\ge4\)

\(\dfrac{4}{b+1}+b\ge3\)(2)

Từ (1),(2) ta có:\(a+\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\ge3\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-b=\dfrac{4}{\left(a-b\right)\left(b+1\right)^2}\\b+1=\dfrac{4}{b+1}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)

2. Ta có\(\dfrac{2a^3+1}{4b\left(a-b\right)}\ge3\)

\(\Leftrightarrow2a^3+1\ge12ab-12b^2\)

\(\Leftrightarrow2a^3+1-12ab+12b^2\ge0\)

\(\Leftrightarrow2a^3-3a^2+1+3\left(a-2b\right)^2\ge0\)

\(\Leftrightarrow\left(2a+1\right)\left(a-1\right)^2+3\left(a-2b\right)^2\ge0\)(luôn đúng)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a-1=0\\a-2b=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=\dfrac{1}{2}\end{matrix}\right.\)