Proof of ##\sqrt[n]{n!}## ≤ ##\frac{n+1}{2}##

[haruspex]

he indicates that no justification is needed

Which strongly suggests he has been unable to do so.
Note that it is not true if we change the 2 in each denominator to e.

 

[chwala]

haruspex i am checking that out, on the side i find maths induction really really interesting. I am dedicating two weeks to study it. I would appreciate if i can get more easy to understand pdf. notes on mathematical induction for beginners...

 

[pasmith]

Homework Statement:: Using mathematical induction show that
$\sqrt[n]n!$ ≤$\frac{n+1}{2}$ where $n$ lies in $z^+$
Relevant Equations:: maths induction

$\sqrt[n]n!$ ≤$\frac{n+1}{2}$​

$\frac {1}{n}$ log [n!]= log$\frac{n+1}{2}$​

You could prove "by induction" that the AM-GM inequality applies to \{1, \dots, n\}, from which the result follows immediately...

 

[haruspex]

haruspex i am checking that out, on the side i find maths induction really really interesting. I am dedicating two weeks to study it. I would appreciate if i can get more easy to understand pdf. notes on mathematical induction for beginners...

As you know, one has to have an inductive hypothesis and a starting point.
In the induction problems given to students, these are both usually given.
In research, the hard part with induction can be picking the right hypothesis. You may set out to prove some property P, but to get the inductive step to work you might need to choose a stronger hypothesis.

 

[chwala]

You could prove "by induction" that the AM-GM inequality applies to \{1, \dots, n\}, from which the result follows immediately...

show me how...now that we already have a solution