a.
Với \(n=1\Rightarrow4\ge3+1\) (đúng)
Giả sử đẳng thức đúng với \(n=k\ge1\) hay \(4^k\ge3k+1\)
Ta cần chứng minh nó cũng đúng với n=k+1 hay: \(4^{k+1}\ge3\left(k+1\right)+1\)
Thật vậy, ta có:
\(4^{k+1}=4.4^k\ge4\left(3k+1\right)=12k+4=3\left(k+1\right)+1+9k>3\left(k+1\right)+1\) (đpcm)
b.
Với \(n=1\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}>1\) (đúng)
Giả sử BĐT đúng với \(n=k\) hay: \(\dfrac{1}{k+1}+\dfrac{1}{k+2}+...+\dfrac{1}{3k+1}>1\)
\(\Rightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1-\dfrac{1}{k+1}\)
Ta cần chứng minh nó cũng đúng với n=k+1 hay:
\(\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3\left(k+1\right)+1}>1\)
\(\Leftrightarrow\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+4}>1\)
Thật vạy, ta có:
\(\dfrac{1}{k+2}+\dfrac{1}{k+3}+..+\dfrac{1}{3k+4}\)
\(=\dfrac{1}{k+2}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}\)
\(>1-\dfrac{1}{k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}\) (1)
Mặt khác ta có:
\(\dfrac{1}{3k+2}+\dfrac{1}{3k+4}-\dfrac{2}{3k+3}=\dfrac{2}{\left(3k+2\right)\left(3k+3\right)\left(3k+4\right)}>0\)
\(\Rightarrow\dfrac{1}{3k+2}+\dfrac{1}{3k+4}>\dfrac{2}{3k+3}\)
\(\Rightarrow\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}>\dfrac{3}{3k+3}=\dfrac{1}{k+1}\) (2)
(1);(2) \(\Rightarrow1-\dfrac{1}{k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}>1\) (đpcm)
