Proving an Inequality: (1+a)q < q/(1-a) for a < 1 and Positive Real Numbers

[grissom1988]

Hello to everyone. This is my first time here so I hope I will not cause any unwanted trouble.

Straight to the problem. I have one inequality for which I would like to prove, but I do not know how. The inequality has the following form:-

(1+a)q < q/(1-a), where a < 1 and q can be any positive real number.

Can someone hint me the direction to prove this? So far, I tried to tackle the inequality by substituting a with different real number smaller than 1, and there is no discrepancy to the inequality.

 

[SammyS]

Hello to everyone. This is my first time here so I hope I will not cause any unwanted trouble.

Straight to the problem. I have one inequality for which I would like to prove, but I do not know how. The inequality has the following form:-

(1+a)q < q/(1-a), where a < 1 and q can be any positive real number.

Can someone hint me the direction to prove this? So far, I tried to tackle the inequality by substituting a with different real number smaller than 1, and there is no discrepancy to the inequality.

If a < 1, then 1-a > 0, in other words 1-a is positive, so you can multiply both sides of the inequality by 1-a.

 

[Ray Vickson]

Hello to everyone. This is my first time here so I hope I will not cause any unwanted trouble.

Straight to the problem. I have one inequality for which I would like to prove, but I do not know how. The inequality has the following form:-

(1+a)q < q/(1-a), where a < 1 and q can be any positive real number.

Can someone hint me the direction to prove this? So far, I tried to tackle the inequality by substituting a with different real number smaller than 1, and there is no discrepancy to the inequality.

Do you really need to keep the 'q'? In other words, as long as q > 0 can you just replace it by q = 1 and still have a true inequality?

RGV

 

[grissom1988]

Hello to everyone. This is my first time here so I hope I will not cause any unwanted trouble.

Straight to the problem. I have one inequality for which I would like to prove, but I do not know how. The inequality has the following form:-

(1+a)q < q/(1-a), where a < 1 and q can be any positive real number.

Can someone hint me the direction to prove this? So far, I tried to tackle the inequality by substituting a with different real number smaller than 1, and there is no discrepancy to the inequality.

I am sorry because I did not put any solid calculation to my question. However, I found a solution to the inequality:-

Originally,

q = q

If I introduce (1-a^2) at the LHS,

(1-a^2)q < q, since a < 1

Also, (1-a^2) = (1+a)(1-a),

Hence (1+a)q < q/(1-a)

Is that a solid proof?

P/s: The q is an important quantity which cannot be deleted. Sorry that I did not mention it earlier.

 

[Ray Vickson]

I am sorry because I did not put any solid calculation to my question. However, I found a solution to the inequality:-

Originally,

q = q

If I introduce (1-a^2) at the LHS,

(1-a^2)q < q, since a < 1

Also, (1-a^2) = (1+a)(1-a),

Hence (1+a)q < q/(1-a)

Is that a solid proof?

P/s: The q is an important quantity which cannot be deleted. Sorry that I did not mention it earlier.

You are getting the idea, but the proof still has holes. Here is why. You start with the statement you want to prove, then introduce some manipulations and end up with another statement that you know to be true. In other words, logically you have (Result) implies (some true statement). What you really need to do is go the other way: (some true statement) implies (Result). This is important, because it is possible to have a false statement imply a true one.

Also: your statement that it is important to keep q is wrong. For example, if q = 7 we want to show that 7(1+a) < 7/(1-a). Would it then not also be true that 5(1+a) < 5/(1-a), or 0.001(1+a) < 0.001/(1-a), or 1(1+a) < 1/(1-a)? Do you really think that the exact value of q contributes to the truth of the inequality? Of course, q must be positive.

RGV