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YONG
tthsqe wrote: (x1)/(x+1) is increasing for x>1 "(a/b)^q > 1" Is this really true? Suppose q = 0. (a/b)^q = 1. Suppose q = 1. (a/b)^q = b/a < 1. 

19 Nov 2015, 04:59 

tthsqe
oops  please change my statement to
(x1)/(x+1) is increasing for x>0 (a/b)^p > (a/b)^q > 0 by hypothesis it is also obvious that (x1)/(x+1) = 12/(x+1) is increasing for x>0 without doing any analysis. 

19 Nov 2015, 05:23 

MHajduk
Below I present my solution of the aforementioned math problem. We prove the given inequality showing stepbystep its equivalence with the condition a > b.
The number of the condition (placed near the "if and only if" connective) used in the particular transition signalizes the importance of the condition for assuring the current step is correct. 

19 Nov 2015, 19:40 

revolution
MHajduk wrote: Below I present my solution of the aforementioned math problem. We prove the given inequality showing stepbystep its equivalence with the condition a > b. 

20 Nov 2015, 01:23 

MHajduk
revolution wrote: Perhaps I misunderstood something but how do you prove that you can eliminate the negative roots? We should all the time bear in mind that we are in the "world" of the positive numbers it's a consequence of the first condition. Or maybe you meant something else? 

20 Nov 2015, 16:13 

revolution
I was thinking that for example sqrt(4) = +/ 2.
You have (pq)root(some positive value). But for integral (pq) there can be both negative and positive solutions. 

20 Nov 2015, 20:43 

MHajduk


20 Nov 2015, 21:17 

revolution
Let's make a=2 and (pq)=2.
(pq)root(a^(pq)) = sqrt(4) = +/ 2. I think you can't run away from the negative result by some algebraic manipulations. Otherwise, from your description above, there would never be any negative roots, ever. 

20 Nov 2015, 21:55 

MHajduk


20 Nov 2015, 23:54 

revolution
I was taught: When taking roots we have to account for both negative and positive roots. The mathematics doesn't know that we only want positive roots, it still gives the valid result of negative roots. We can choose to ignore negative roots but then that makes any proofs based upon it invalid or incomplete.
I still claim 2 is a perfectly valid solution to the square root of 4. Last edited by revolution on 21 Nov 2015, 05:50; edited 1 time in total 

21 Nov 2015, 01:47 

TmX
I guess this is a sort of convenience/convention.
When asked what are the solutions of x^2 = 25, you will be expected to answer both 5 and 5. But when asked what is the square root of 25, just answer 5. Maybe most people prefer injective function 

21 Nov 2015, 05:47 

YONG
Let's try another route.
TRUE <=> a > b <=> ln a > ln b [since a & b are +ve and ln is monotonic increasing] <=> (pq) ln a > (pq) ln b [since (pq) > 0] <=> ln [a^(pq)] > ln [b^(pq)] [by properties of logarithm] <=> a^(pq) > b^(pq) [since ln is monotonic increasing] 

21 Nov 2015, 06:04 

revolution
I think the trick is simply to realise that (pq) > 0. And simplify to an equivalent a^c > b^c where c=(pq) and a>b>0.
YONG's solution also works for me. 

21 Nov 2015, 07:25 

MHajduk
revolution wrote: I was taught: When taking roots we have to account for both negative and positive roots. The mathematics doesn't know that we only want positive roots, it still gives the valid result of negative roots. We can choose to ignore negative roots but then that makes any proofs based upon it invalid or incomplete. a^(pq) > b^(pq) <=> a > b
revolution wrote: I still claim 2 is a perfectly valid solution to the square root of 4. If we are talking about solutions of the relation y^2 = x in R^2 then we have to accept both pairs (x, sqrt(x)) and (x, sqrt(x)). The widely accepted practice is to treat the sign of the square root as a symbol of a function (as long as we don't move to the field of the complex numbers where we have to accept all n roots of the nth root of the given numbers, both real and complex). 

21 Nov 2015, 16:55 

MHajduk
A stained glass effect test. This was something truly laborious from the technical point of view (a lot of work with lighting).


27 Nov 2015, 00:14 

tthsqe
I would say taylor's inequality for sqrt(x) about x=1. This works for both sides of the inequality if you dehomogenize it both ways.


28 Nov 2015, 03:41 

Xorpd!
I've never heard of Taylor's inequality, but here is an approach which seems to work:
Let f(a,b) = 4a(a+b)8√(a³b)(ab)² = 3a²+6abb²8√(a³b) Then on the left boundary of the domain, a = b, f(a,a) = 0. ∂f/∂a = 6a+6b12√(ab) = 6(√a√b)² > 0 for a > b, so the function increases from left to right and so is nonnegative over its domain. It's easy to get from there to the left inequality. Let g(a,b) = (ab)²4b(a+b)+8√(ab³) = a²6ab3b²+8√(ab³) On the upper boundary of the domain, a = b, g(b,b) = 0. ∂g/∂b = 6a6b+12√(ab) = 6(√a√b)² < 0 for a > b, so the function increases from top to bottom and so is nonnegative over its domain. This yields the right inequality. Edit: corrected 2 typos. 

28 Nov 2015, 05:15 

MHajduk
tthsqe wrote: I would say taylor's inequality for sqrt(x) about x=1. This works for both sides of the inequality if you dehomogenize it both ways. 

29 Nov 2015, 17:49 

MHajduk
The approach presented by Xorpd! gives me a new perspective of solving inequalities. I dared to formulate his solution in a more descriptive manner (as I understood it):


29 Nov 2015, 17:52 

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