a.
Ta chứng minh bằng quy nạp:
Với \(n=1\Rightarrow a_1>\dfrac{1}{2}-\dfrac{1}{2.1}=0\) (đúng)
Giả sử BĐT đúng với \(n=k\) hay: \(a_k>\dfrac{1}{2}-\dfrac{1}{2k}\)
Ta cần chứng minh \(a_{k+1}>\dfrac{1}{2}-\dfrac{1}{2\left(k+1\right)}\)
Thật vậy, ta có:
\(\dfrac{1}{4}\le a_{k+1}\left(1-a_k\right)< a_{k+1}\left[1-\left(\dfrac{1}{2}-\dfrac{1}{2k}\right)\right]\)
\(\Rightarrow\dfrac{1}{4}< a_{k+1}\left(\dfrac{1}{2}+\dfrac{1}{2k}\right)\)
\(\Rightarrow a_{k+1}>\dfrac{k}{2\left(k+1\right)}=\dfrac{1}{2}-\dfrac{1}{2\left(k+1\right)}\)
b.
\(\dfrac{1}{4}\le a_{n+1}\left(1-a_n\right)\)
\(\Rightarrow\dfrac{1}{2}\le\sqrt{a_{n+1}\left(1-a_n\right)}\le\dfrac{1}{2}\left(a_{n+1}+1-a_n\right)\)
\(\Rightarrow a_{n+1}-a_n\ge0\)
Dãy đã cho là dãy ko giảm
