An inequality: (sqrt(1-a)-sqrt(1-b))^2 >= (sqrt(1-a/c)-sqrt(1-b/c))(sqrt(1-ac)-sqrt(1-b*c))

[m.alimohammadi.stat]

Dear All

Can you prove the following inequality when 0<a<b<1 for any c>=1 ?


(sqrt(1-a)-sqrt(1-b))^2 >= (sqrt(1-a/c)-sqrt(1-b/c))*(sqrt(1-a*c)-sqrt(1-b*c)).

Thanks in advance

 

[Dr.Peterson]

Dear All

Can you prove the following inequality when 0<a<b<1 for any c>=1 ?


(sqrt(1-a)-sqrt(1-b))^2 >= (sqrt(1-a/c)-sqrt(1-b/c))*(sqrt(1-a*c)-sqrt(1-b*c)).

Thanks in advance

First, let's check your goal. Is it to prove this? [math]\left(\sqrt{1-a}-\sqrt{1-b}\right)^2 \ge \left(\sqrt{1-\frac{a}{c}}-\sqrt{1-\frac{b}{c}}\right)*\left(\sqrt{1-ac}-\sqrt{1-bc}\right)[/math]
Next, please keep in mind that what we do here is to help you solve your problem, not just to do it ourselves; so we need to see what you have tried, or at least what you know that you can use. (I haven't yet tried to prove this, so I can't answer your question by saying yes or no.)

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[m.alimohammadi.stat]

Dear Dr. Peterson

Thanks for your attention.
Yes, I meant that.
However, I have a proof for a fewer case:
(sqrt(1-a))^2 >= (sqrt(1-a/c))*(sqrt(1-a*c)).
It is based on the concept of log concavity and we use the function f(x)=sqrt(1-x) for it.
Now, my problem is in the general case mentioned in the previous post and I need a proof without the concept of log concavity. Indeed, I need a proof based on some well-known mathematical inequalities (if possible).

Warm regards.

 

[lookagain]

Can you prove the following inequality when 0<a<b<1 for any c>=1 ?
If

(sqrt(1-a)-sqrt(1-b))^2 >= (sqrt(1-a/c)-sqrt(1-b/c))*(sqrt(1-a*c)-sqrt(1-b*c)).

From what I can see in your restrictive horizontal spacing, you have left out needed grouping symbols. In Latex, this appears what you intended:

\(\displaystyle \bigg(\sqrt{1 - a} - \sqrt{1 - b}\bigg)^2 \ge \bigg(\sqrt{\dfrac{1 - a}{c}} - \sqrt{\dfrac{1- b}{c}}\bigg)\bigg(\sqrt{(1 - a)c} - \sqrt{(1 - b)c}\bigg)\)

Try expanding the right-hand side. If you want to look at a specific example (not part
of a proof), you can let a = 3/4 and b = 15/16 for convenient square roots upon
substitution. Also, you could let c = 4 so that it is not a trivial 1 in the evaluation.

 

[Dr.Peterson]

Can you prove the following inequality when 0<a<b<1 for any c>=1 ?

I can now say definitively, my answer is NO.

I tried evaluating both sides for various values (including graphing them with c varying, and changing the radicands by adding parentheses), and if the inequality had been [imath]\le[/imath] it looks like it would be correct, but not as written. (I didn't go as far as @lookagain in modifying the expressions, because that makes it almost trivial, and not really an inequality.)

Something we forgot to ask for before is an explanation of why you ask this. Is it part of a bigger problem?

 

[m.alimohammadi.stat]

I can now say definitively, my answer is NO.

I tried evaluating both sides for various values (including graphing them with c varying, and changing the radicands by adding parentheses), and if the inequality had been [imath]\le[/imath] it looks like it would be correct, but not as written. (I didn't go as far as @lookagain in modifying the expressions, because that makes it almost trivial, and not really an inequality.)

Something we forgot to ask for before is an explanation of why you ask this. Is it part of a bigger problem?

Dear Dr. Peterson

I replied to your first message at the same time, but it was published one day late.
If you are sure that the current inequality does not hold, then it is of no use to me.
Also this was to find an example for a theorem that according to your answer I will use the same inequality in my previous message with fewer functions.
Thank you again for your comments and your time.

Warm wishes.

 

[Dr.Peterson]

Dear Dr. Peterson

I replied to your first message at the same time, but it was published one day late.
If you are sure that the current inequality does not hold, then it is of no use to me.
Also this was to find an example for a theorem that according to your answer I will use the same inequality in my previous message with fewer functions.
Thank you again for your comments and your time.

Warm wishes.

Let's just take an example, so you can be sure. Suppose [imath]a=0.1,b=0.5,c=1.25[/imath].

Then

[math]\left(\sqrt{1-a}-\sqrt{1-b}\right)^2=\left(\sqrt{1-0.1}-\sqrt{1-0.5}\right)^2=\left(\sqrt{0.9}-\sqrt{0.5}\right)^2=0.058359[/math]
[math]\left(\sqrt{1-\frac{a}{c}}-\sqrt{1-\frac{b}{c}}\right)\left(\sqrt{1-ac}-\sqrt{1-bc}\right)=\left(\sqrt{1-\frac{0.1}{1.25}}-\sqrt{1-\frac{0.5}{1.25}}\right)\left(\sqrt{1-0.1\cdot1.25}-\sqrt{1-0.5\cdot1.25}\right)=\left(\sqrt{0.92}-\sqrt{0.6}\right)\left(\sqrt{0.875}-\sqrt{0.375}\right)=0.0596237[/math]
The former is less than the latter. Here is a graph comparing the sides, taking those values of a and b, as c varies:

​

LHS = red dots [imath]\le[/imath] RHS = green.