Assume (n!)1/n <= (n+1)/2 for all n>=1 .................(1)
<=> n! <= ( (n+1)/2 )n for all n>=1
<=> 2n * n! <= (n+1)n for all n>=1
So, if we can prove 2n * n! <= (n+1)n is true for all integer n >=1 then the inequality 1 will also be proved.
Statement P(n) : 2n * n! <= (n+1)n
We have to prove P(n) is true for all integer n>=1
Base case : P(1) : 2<=2
So, P(1) is true
Inductive Hypothesis : We assume P(k) is true i.e 2k*k! <= (k+1)k for some integer k>=1 ................(2)
Inductive Step : Assuming the statement (2) we have to prove P(k+1) is true.
P(k+1) : 2k+1 * (k+1)! <= (k+2)(k+1)
From statement (2) we get
2k *k! <= (k+1)k
Multiplying both sides of the inequality with 2(k+1) we get
2(k+1) * (k+1)! <= 2(k+1)k+1 <= (k+2)k+1 ( Since 2nn <= (n+1)n (THINK) )
Hence 2(k+1) * (k+1)! <= (k+2)k+1 ( Proved )
That means whenever P(k) is true P(k+1) is also true.
So, P(n) is true for all integer n>=1
Hence the inequality (n!)1/n <= (n+1)/2 is true for all integer n>=1
So, Option A is the correct answer.