Inequality involving two unknowns and factorials

In summary, the conversation discusses proving the inequality n!/r! >= r^(n-r) with n and r as natural numbers and n>=r. The speaker considers different approaches, including using induction, and ultimately chooses to prove it for n=1 and then extend it to all n by showing that each term in the LHS is greater than r^(n-r). They also mention the simplicity of another person's approach.
  • #1
Rosey24
12
0
I need to prove the following:
n!/r! >= r^(n-r)

With r and n as natural numbers and n>=r
I know the LHS will end up being (n-r) terms long as the first r! will cancel out of n! (n>=r), but as they're both unknown, I just left it as
1*2*3*...*(n-2)*(n-1)*n

1*2*3*...*(r-2)*(r-1)*r

and I looked at the RHS as r^n/r^r, but I'm not sure how to work with these two sides. :uhh:

Any tips on different ways to approach this would be greatly appreciated.

Thanks!


Update: Is it enough to do the following, using induction:
Prove that for n=1, it works as 1 >= 1
and then, assuming the inequality holds for n, try n+1:
(n+1)!/r! >= r^[(n+1)-r]
which is just
n!(n+1)/r! >= r^(n-r)*r
and, as n>=r, (n+1)>=r, so, by induction, it's holds for all n.
 
Last edited:
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  • #2
That's the way I would do it! For a moment I thought- "He's assuming r= 1 and he can't do that", but you can. Since r is a natural number, less than or equal to n, if n= 1 then r= 1.
 
  • #3
LHS= (r+1)*(r+2)...*n , there are n-r terms and each term is greater than r . Hence, the LHS is greater than r^(n-r), right?
 
  • #4
Ahh I like the way this new guy thinks :) Much simpler than induction, original and just plain :) which is good lol.
 
  • #5
Thanks to everyone who responded!

I appreciate the simplicity of your response, huyen_vyvy. One small note: it's a she who assumed r=1, not a he.
 

1. How do you solve an inequality involving two unknowns and factorials?

To solve an inequality involving two unknowns and factorials, first simplify the factorial expressions as much as possible. Then, isolate one of the unknown variables on one side of the inequality and the other on the other side. Finally, use algebraic manipulation and/or trial and error to find the values of the variables that satisfy the inequality.

2. What are the key properties of factorials that are important in solving inequalities?

The key properties of factorials that are important in solving inequalities are the fact that factorials grow very quickly, the fact that they are always positive, and the fact that they are not defined for negative integers.

3. Can inequalities involving factorials have more than one solution?

Yes, inequalities involving factorials can have multiple solutions. This is because factorials are continuous functions, meaning that for every point on the number line, there is a corresponding value for the factorial. Therefore, there can be multiple values that satisfy an inequality involving factorials.

4. How can I check my solution for an inequality involving factorials?

You can check your solution for an inequality involving factorials by plugging in the values of the variables into the original inequality and seeing if it is true. Additionally, you can use a graphing calculator or software to graph the inequality and see if your solution falls within the shaded region.

5. Are there any special cases or restrictions when solving inequalities involving factorials?

Yes, there are a few special cases and restrictions to keep in mind when solving inequalities involving factorials. For example, factorials are not defined for negative integers, so any solutions that make the factorial expression negative are not valid. Additionally, some inequalities may have no solution or an infinite number of solutions, so it is important to carefully analyze the inequality before solving.

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