Induction proof of an inequality

AI Thread Summary
The discussion focuses on proving the inequality n! ≤ n^n for all integers n ≥ 1 using mathematical induction. The base case is established with 1! ≤ 1^1, confirming the statement holds for n=1. The inductive hypothesis assumes k! ≤ k^k is true, leading to the need to prove (k+1)! ≤ (k+1)^(k+1). Participants suggest rewriting (k+1)! in terms of k and k^k to facilitate the proof, indicating that the right-hand side of the inequality needs correction. The conversation emphasizes the importance of correctly manipulating the terms to complete the induction step.
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Homework Statement



for all integers n>=1, n! <= n^n

Homework Equations





The Attempt at a Solution



Base case: (1)! <= (1)^(1) 1=1 check
Inductive hypothesis: suppose k!<=k^k
P(k+1): (k+1)! <= (k+1)^(k+1)

From here on out I get very confused. Any help would be appreciated!
 
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Write (k+1)! \le (k+1)^{k+1} in terms of k and k^k.
 
so it would be k!(k+1) <= (k+1)^k + (k+1) ?
 
The right hand side is incorrect, but you're on the right track.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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