MHB Proving the Sum of a Sequence Equals n Using Induction

[joypav]

Okay, so I need to prove this. I thought I would be using induction, right?

First we can consider the base case, which is simple.

Next we have to do the induction step.

I think we consider one case where each a=1. Then we have 1<=1.

Then consider that they are not all 1. I can't remember what trick to use. Is it that there must exist at least one a such that a<1? Because if they're all larger than 1 then the sum can't be n. And then there must also exist an a>1? Because if they're all less than 1 then the sum can't be n.

 

[joypav]

Also, just to clear it up, I don't want to use AGM to prove it. The point is to prove this without AGM.

 

[Euge]

Hi joypav,

Using the fact that $\log x \le x - 1$ for all $x > 0$, you can prove directly that $\log(a_1\cdots a_n) \le 0$, which is equivalent to $a_1\cdots a_n \le 1$.

 

[joypav]

Hi joypav,

Using the fact that $\log x \le x - 1$ for all $x > 0$, you can prove directly that $\log(a_1\cdots a_n) \le 0$, which is equivalent to $a_1\cdots a_n \le 1$.

My professor actually provided me with a much different solution. This one is simpler though! Thanks

 

[Euge]

Out of curiosity, what was the solution your professor provided?

 

[joypav]

Out of curiosity, what was the solution your professor provided?

It was a proof by induction.
The typical basis step.
Then the induction step by considering that there must be at least one value greater than one and at least one value less than one.

If you're interested I can type it out.

 

[Euge]

There's no need. I was just curious what he did.