Prove a+b < 1+ab for a, b > 1

[armolinasf]

Homework Statement

Prove that if a,b > 1, then a+b < 1+ab

The Attempt at a Solution

Just want to know if this makes sense:

first let a+b < 1+ab become 1<(1+ab)/(a+b) ==> 0<(1+ab-(a+b))/(a+b).

Factoring the numerator: 0<(1-a+ab-b)/(a+b) ==> 0<(1-b)+a(b-1)/(a+b)

So the next step would be to figure out where the numerator is greater than zero, since that is equivalent to our original inequality (do we just ignore the denominator since its undefined only in relation to variables?).

Solving the numerator for b would be: 0<(1-b)+a(b-1) ==> -(1-b)<a(b-1) ==> 1<a

Would this be an acceptable proof?

 

[CompuChip]

It would be, if you wrote it down the other way around.
So start with: "suppose that a, b > 1. Then 1 < a, therefore -(1 - b) < a(b - 1). So 0 < (1 - b) + a (b - 1) " ... etc.

If you can do that an each step is still valid, then the proof is correct.

 

[hunt_mat]

What about examining the following:
$$a+b-(1+ab)$$
This can be factorised:
$$a+b-1-ab=a(1-b)+b-1=(b-1)(1-a)$$

 

[zgozvrm]

So the next step would be to figure out where the numerator is greater than zero...

Actually, you need to show that the numerator IS greater than 0 for all a,b (not where it is greater).

do we just ignore the denominator since its undefined only in relation to variables?

Yes and no.

Yes. You can ignore the denominator since we know a,b > 1 then the denominator (a+b) must also be greater than 1

No. The denominator is not undefined at all, since it can never be equal to 0.

Solving the numerator for b would be: 0<(1-b)+a(b-1) ==> -(1-b)<a(b-1) ==> 1<a

Would this be an acceptable proof?

No. This is incomplete, it only shows that a>1; what about b>1?

 

[zgozvrm]

first let a+b < 1+ab become 1<(1+ab)/(a+b) ==> 0<(1+ab-(a+b))/(a+b).

Factoring the numerator: 0<(1-a+ab-b)/(a+b) ==> 0<(1-b)+a(b-1)/(a+b)

...

Solving the numerator for b would be: 0<(1-b)+a(b-1) ==> -(1-b)<a(b-1) ==> 1<a

You could save yourself some time (and, at the same time, avoid the issue of the denominator) by simply subtracting (a+b) from the original inequality:

a+b < 1 + ab
0 < 1 + ab - a - b
0 < (1 - b) + (ab - a)
0 < (1 - b) + a(b - 1)
-(1 - b) < a(b - 1)
(b - 1) < a(b - 1)
1 < a

 

[hunt_mat]

You could save yourself some time (and, at the same time, avoid the issue of the denominator) by simply subtracting (a+b) from the original inequality:

a+b < 1 + ab
0 < 1 + ab - a - b
0 < (1 - b) + (ab - a)
0 < (1 - b) + a(b - 1)
-(1 - b) < a(b - 1)
(b - 1) < a(b - 1)
1 < a

What this not what I already said?

 

[zgozvrm]

What this not what I already said?

Essentially, yes. I just made it more clear how the result the OP came up with can be obtained by showing the steps from beginning to end. (At the same time answering the question concerning the denominator).

 

[zgozvrm]

I would look at it in reverse: Given that you have a>1 and b>1, can you show that a + b < 1 + ab?

Starting out, we already know that a>1 and b>1
Therefore, we know that a-1 > 0 (subtracting 1 from both sides)
(a-1) is therefore non-zero and positive.
We can then multiply both sides of b>1 by (a-1), giving us b(a-1) > (a-1)
Multiplying out, we get ab - b > a - 1
Rearranging, we get ab + 1 > a + b
Which is the same as a+b < 1 + ab thus proving the given statement.

 

[armolinasf]

I never thought of looking at the problem that way zgozvrm. Thanks for all the advice

 

[eczeno]

i feel the need to stress, as compuchip pointed out, and zgozvrm demonstrated, to prove a conditional (if A, then B) you can assume A and show that B follows, but not the other way around. Assuming B and showing A (which is what the original post tried to do) would not be a proof of 'If A, then B'.