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A problem about logarithmic inequality

  • #1

Homework Statement


If a,b,c are positive real numbers such that ##{loga}/(b-c) = {logb}/(c-a)={logc}/(a-b)## then prove that
(a) ##a^{b+c} + b^{c+a} + c^{a+b} >= 3##
(b) ##a^a + b^b + c^c >=3##

Homework Equations


A.M ##>=## G.M

The Attempt at a Solution


Using the above inequation, I am able to get (b) but I have no idea to do (a) keeping in mind the exponents as a sum whereas in the question the denominators are as a difference.
 

Answers and Replies

  • #2
Charles Link
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It doesn't look like a one or two step problem, but I did get the result from the set of logarithmic equations (by combining a couple of them) that abc=1. It does appear when a=b=c=1 that the equality holds. For any other case, you then need to show that the "greater than" sign applies.
 
  • #3
Ray Vickson
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It doesn't look like a one or two step problem, but I did get the result from the set of logarithmic equations (by combining a couple of them) that abc=1. It does appear when a=b=c=1 that the equality holds. For any other case, you then need to show that the "greater than" sign applies.
It is not allowed to have ##a = b = 1## (or even ##a = b = c \neq 1##) because that would involve division by zero in the constraint
[tex] \log(a)/(b-c) = \log(b)/(c-a) = \log(c)/(a-b) [/tex]
Even trying to take limits as ##a \to 1##, ##b \to 1## and ##c \to 1## will not really work, because depending on exactly how those limits are taken, the quantities involved in the constraint can still fail to exist, or can exist but have more-or-less arbitrary values.
 
  • #4
Charles Link
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It is not allowed to have ##a = b = 1## (or even ##a = b = c \neq 1##) because that would involve division by zero in the constraint
[tex] \log(a)/(b-c) = \log(b)/(c-a) = \log(c)/(a-b) [/tex]
Even trying to take limits as ##a \to 1##, ##b \to 1## and ##c \to 1## will not really work, because depending on exactly how those limits are taken, the quantities involved in the constraint can still fail to exist, or can exist but have more-or-less arbitrary values.
The use of the "logs" to state the problem appears to be largely algebraic. After processing these equations, the result emerges that abc=1 and the logs wind up being removed. The problem could actually begin with, given a,b, c positive real, such that abc=1, prove the two inequalities. I think you will find that the equality holds only for the case a=b=c=1. I did not do the proof in its entirety, partly because on a homework problem we are not supposed to present the whole solution, but I think the only case where the equal sign would apply is for a=b=c=1. (I agree, this puts a zero in the denominators of the original statement of the problem.)
 
  • #5
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We may assume that either ##0 < a < b < c## or ##0 < a < c < b##. As @Ray Vickson pointed out they cannot be equal. Neither can one of them equal one for then all of them would be equal one, especially they would be equal.
The rest is simply about cases.
Assume now ##0<a<b<c##. If ##c < 1## then ##\frac{\log a}{b-c} > 0## and ##\frac{\log b}{c-a} < 0##, a contradiction.
In the same way ##b < 1## leads to ##\log a < \log b < 0 ## and ##\frac{\log a}{b-c} > 0## and ##\frac{\log b}{c-a} < 0##, a contradiction.
If ## a < 1 ## then ##\frac{\log a}{b-c} > 0## and ##\frac{\log c}{a-b} < 0##, a contradiction.
This means ##a, b, c > 1## and both assertions are true. (The second case ##0 < a < c < b## should be analogue.)

The principle behind it is the following: The assumption is symmetric in ##(a,b,c),## i.e. with positive signum. On the other side there has to be a change in signs if not all of them are positive factors which means greater than one for the nominators.
 
  • #6
Ray Vickson
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The use of the "logs" to state the problem appears to be largely algebraic. After processing these equations, the result emerges that abc=1 and the logs wind up being removed. The problem could actually begin with, given a,b, c positive real, such that abc=1, prove the two inequalities. I think you will find that the equality holds only for the case a=b=c=1. I did not do the proof in its entirety, partly because on a homework problem we are not supposed to present the whole solution, but I think the only case where the equal sign would apply is for a=b=c=1. (I agree, this puts a zero in the denominators of the original statement of the problem.)
I suspect that as long as ##a,b,c## are valid (no division by 0) the inequalities to be shown will be strict.
 
  • #7
Charles Link
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I will show the algebra that gives the result ## abc=1 ##. Removing the log factors on both sides, ## a^{c-a}=b^{b-c} ## and ## a^{a-b}=c^{b-c} ##. (also ## b^{a-b}=c^{c-a} ##). Taking the first two equations and multiplying left sides and right sides, ## a^{c-b}=(bc)^{b-c} ## so that ## a^{c-b}=1/(bc)^{c-b} ##. Thereby ## (abc)^{c-b}=1 ##. I think it can be concluded that ## abc=1 ##. (You can write similar equations by combining the other pairs of equations also with the result that ## abc=1 ##).The way the problem is written with the factors in the denominators places additional restrictions, but otherwise, it would be possible to have ## a=b=c=1 ##. To proceed from here, I would let ## c=1/(ab) ## and write the inequalities as functions of ## a ## and ## b##. There are basically two cases that need to be tested, ## ab>1 ## and ## ab<1 ##. The way the problem is written yes, if one of the three letters is equal to 1, then all 3 must equal one (log 1=0, etc), but that makes the denominators all zero. Instead of using logs, they could simply have presented the condition ## abc=1 ##. (with a,b, c positive real). In that case, the problem would be less restrictive, but the same inequalities would apply.
 
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  • #8
Charles Link
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We may assume that either ##0 < a < b < c## or ##0 < a < c < b##. As @Ray Vickson pointed out they cannot be equal. Neither can one of them equal one for then all of them would be equal one, especially they would be equal.
The rest is simply about cases.
Assume now ##0<a<b<c##. If ##c < 1## then ##\frac{\log a}{b-c} > 0## and ##\frac{\log b}{c-a} < 0##, a contradiction.
In the same way ##b < 1## leads to ##\log a < \log b < 0 ## and ##\frac{\log a}{b-c} > 0## and ##\frac{\log b}{c-a} < 0##, a contradiction.
If ## a < 1 ## then ##\frac{\log a}{b-c} > 0## and ##\frac{\log c}{a-b} < 0##, a contradiction.
This means ##a, b, c > 1## and both assertions are true. (The second case ##0 < a < c < b## should be analogue.)

The principle behind it is the following: The assumption is symmetric in ##(a,b,c),## i.e. with positive signum. On the other side there has to be a change in signs if not all of them are positive factors which means greater than one for the nominators.
If ## a<b<c ## then ## c>1 ##. But ## b ## could be greater than one or less than one, and ## a<1 ##. Please double-check your logic.
 
  • #9
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If ## a<b<c ## then ## c>1 ##. But ## b ## could be greater than one or less than one, and ## a<1 ##. Please double-check your logic.
The logic is ok, but you are right: there is no solution at all.
Three positive real numbers can be ordered. They cannot be pairwise equal for this meant a division by zero.
Therefore there has to be at least one change of sign in the denominators: ##b-c, c-a, a-b##. To be equal this means there is a change in the sign of the nominators. However, the nominators are somehow linearly ordered, e.g. ##0 < a < b < c,## and the denominators are cyclic. Since the logarithm is strictly monotone, the two orderings cannot be true at the same time.

In your example with ##0 < a < b < c## we get ##\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{-}=\frac{\log b}{+}=\frac{\log c}{-}##. On the other hand ##\log a < \log b < \log c## and the nominator in the middle cannot have a different sign than the two left and right.

The assumption is always wrong.
 
  • #10
Charles Link
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The logic is ok, but you are right: there is no solution at all.
Three positive real numbers can be ordered. They cannot be pairwise equal for this meant a division by zero.
Therefore there has to be at least one change of sign in the denominators: ##b-c, c-a, a-b##. To be equal this means there is a change in the sign of the nominators. However, the nominators are somehow linearly ordered, e.g. ##0 < a < b < c,## and the denominators are cyclic. Since the logarithm is strictly monotone, the two orderings cannot be true at the same time.

In your example with ##0 < a < b < c## we get ##\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{-}=\frac{\log b}{+}=\frac{\log c}{-}##. On the other hand ##\log a < \log b < \log c## and the nominator in the middle cannot have a different sign than the two left and right.

The assumption is always wrong.
A very good observation. Yes, the problem statement is clearly inconsistent. (unless we are missing the obvious). I agree with your assessment.
 
  • #11
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Well that rules out a<b<c, but what about a<c<b?
$$\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{+}=\frac{\log b}{+}=\frac{\log c}{-}$$
Clearly log(c) has a different sign than the others, so 0<a<b<1<c. The equalities are cyclic, but swapping the parity makes a difference.

Edit: no, doesn't work either. c-a is the largest denominator, so log(b) has to be the largest numerator in magnitude, but log(a) is larger.
Both cases are excluded, but the second one is a bit more subtle.
 
  • #12
Ray Vickson
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Well that rules out a<b<c, but what about a<c<b?
$$\frac{\log a}{b-c} = \frac{\log b}{c-a}=\frac{\log c}{a-b}=\frac{\log a}{+}=\frac{\log b}{+}=\frac{\log c}{-}$$
Clearly log(c) has a different sign than the others, so 0<a<b<1<c. The equalities are cyclic, but swapping the parity makes a difference.

Edit: no, doesn't work either. c-a is the largest denominator, so log(b) has to be the largest numerator in magnitude, but log(a) is larger.
Both cases are excluded, but the second one is a bit more subtle.
There is no need to check other cases. The three numbers must all be different. Let's call the smallest of the three ##a##, the largest of the three ##c## and the one in the middle ##b##. In the original problem there were no characteristics distinguishing ##a, b## or ##c##, so there were no a priori ways to distinguish between them. Basically, it is just saying "without loss of generality, assume ##a < b < c##", and we do that type of thing all the time.
 
  • #13
Charles Link
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@fresh_42 made a very good observation. I've looked hard to try to find an exception to this that would make it work, but it seems the problem in the way it is formulated is inconsistent. Writing the expressions without the logs: ## a^{1/(b-c)}=b^{1/(c-a)}=c^{1/(a-b)} ## apparently it is impossible for all three terms to be greater than 1 or all three terms to be less than one. If two of them are greater than one, the third must be less than one, etc...
 
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  • #14
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Okay, checking that in more detail, swapping a and b doesn't change the equalities, but that is not as obvious as an a->b->c->a rotation.
 

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