LoL OMG how did I end up here? Still at least I know it is possible to do equations on computers. Can computers give you answers to those equations???? *probably a stupid question*
Notice that H1×H2={(g1,g2)∈G1×G2∣(g1,e)∈K,(e,g2)∈K} As K is a group, it follows that H1≤G1, H2≤G2 Moreover, if (g1,g2)∈H1×H2, then, obviously, (g1,g2)∈K. It is easy to see that (g1−1,e) and (e,g2−1) are in K, and therefore (g1−1,g2−1)∈H1×H2. Furthermore, (g1,g2),(g3,g4)∈H1×H2 implies (g1g3,g2g4)∈H1×H2. Trivially, (e,e)∈H1×H2. We can conclude that H1×H2 is a subgroup of K. Suppose that there is at least one element (g1,g2)∈K, which is not in H1×H2. The numbers card(G1) and card(G2) are relatively prime. Consequently, we can choose n1,n2 such that n1card(G1)≡1(modcard(G2)) and n2card(G2)≡1(modcard(G1)). Then, (g1n1card(G1),g2n1card(G1))=(e,g2)∈K and (g1n2card(G2),g2n2card(G2))=(g1,e)∈K. Hence, (g1,g2)∈H1×H2, which is clearly a contradiction.
(You'll probably all find this easy, but perhaps its a good one to make you think about initial approach. Plus i made it myself!) Problem 73*
i) Find the number of ordered sets of positive integers x1,x2,x3,...,xn such that i=1∑nxi−i=1∏nxi=n−1 and xi≤m where m is a positive integer such that m≥n. Justify your answer. ii) Extend your result to include all ordered sets of non-negative integers.
No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.
No i deleted it, as I considered how inadequate my knowledge of the subject was having not looked at hyperboilc functions. It just seems to come straight from the line above that, but I'm probably missing something. How did you manage to reply to that, i thought I deleted it in the nick of time.
If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.
If you have a message quoted, I think it posts it regardless of whether the other person deletes it. When you delete a message, it only really removes it from view for the majority of the members of the forum, but it still has a unique post code and can still be quoted - so if someone quotes you before you delete it, it will have all the text you posted as well.
Let a,b,c be positive real numbers. Then we have the following inequality: (b+c)2a+(c+a)2b+(a+b)2c+(a+b)(b+c)(c+a)2(ab+bc+ca)≥2(ab+bc+ca)3(a+b+c).
Let a,b,c be positive real numbers. Then we have the following inequality: (b+c)2a+(c+a)2b+(a+b)2c+(a+b)(b+c)(c+a)2(ab+bc+ca)≥2(ab+bc+ca)3(a+b+c).
Spoiler
Spoiler
Noticing that either side on the equation is symmetrical in terms of the interchangeability of a, b and c, we can assume, W.L.O.G, that 0≤a≤b≤c (*).
Simplifying we get: LHS= (b+c)2a+(c+a)2b+(a+b)2c+(a+b)(c+a)a+(a+b)(b+c)b+(b+c)(c+a)c
As each term is positive and clearly the denominator of each term in less than or equal to (c+c)2=4c2 by (*). Simplifying we get... LHS≥2c2(a+b+c) for ALL a, b and c that satisfy (*). The maximum value of this expression occurs when a=b=c (i.e: a+b+c=3c) so we get LHS≥2c3=2a3 (by the fact a=b=c had to be assumed in order to get the maximum).
Using the same technique, RHS≤2(3a2)3(3c)=2a23c. This clearly occurs when a=b=c also so RHS≤2a3
We have 2a3≤LHS≤2a2a+b+c for all a,b,c. Hence 2a3≤LHS≤2a3 and equality occurs only when a=b=c?!
Edit: I mean, what makes you believe that LHS≥2a3 is always true? You have LHS≥2c2a+b+c for all a,b,c, but does this imply LHS≥max(2c2a+b+c) for all a,b,c? Bear in mind that LHS changes as you change a,b,c.
We have 2a3≤LHS≤2a2a+b+c for all a,b,c. Hence 2a3≤LHS≤2a3 and equality occurs only when a=b=c?!
I haven't got an upper bound for the LHS (I said it was greater that 2a2a+b+c I think you misread!)
Edit: I mean, what makes you believe that LHS≥2a3 is always true? You have LHS≥2c2a+b+c for all a,b,c, but does this imply LHS≥max(2c2a+b+c) for all a,b,c? Bear in mind that LHS changes as you change a,b,c.
Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that... 2a23c≥ LHS ≥2c23a by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!
Hmm I see your point. Well, short of using partial differentiation to justify this is a minimum, would it be sufficient to note that... 2a23c≤ LHS ≥2c23a by (*) and note that as a approaches c, each inequality approaches equality and hence demonstrate that the minimum value occurs at a=b=c? (if this works then the same could be easily done for the RHS). Cheers for the feedback btw, I wasn't sure it worked but I thought I'd give it go!
Your argument proves only that LHS=2a3, when a=b=c. Also, should it not be 2a23c≥LHS?
If we denote f(a,b,c)=LHS, then we have to prove that there are no a,b,c such that 2a3>f(a,b,c)≥2c2a+b+c; in other words, we have to prove that f reaches its minimum when a=b=c.
Partial differentiation would work, but it is quite tough task.
I gave you the upper bound just to make you think that you need to justify your inequalities.
Your argument proves only that LHS=2a3, when a=b=c. Also, should it not be 2a23c≥LHS?
Sorry, I'll change that now
If we denote f(a,b,c)=LHS, then we have to prove that there are no a,b,c such that 2a3>f(a,b,c)≥2c2a+b+c; in other words, we have to prove that f reaches its minimum when a=b=c.
Partial differentiation would work, but it is quite tough task.
Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that... ∂a∂LHS,∂b∂LHS,∂c∂LHS<0∀a,b,c>0 which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.
Well using partial derivatives on the 6-term expression with respect to a, b and c in turn we get that... ∂a∂LHS,∂b∂LHS,∂c∂LHS<0∀a,b,c>0 which implies f(a,b,c) approaches a stationary point as all three variables coalesce to zero. Choosing a few values above and below shows you that this is a minimum point and hence, the minima occurs when and only when a=b=c.
You are right, the points, at which the partial derivatives fail to exist, are good candidates for extremums. Yet, this does not help us at all. I tried several methods to prove your inequalities, including some brute-force techniques such as Lagrange Multipliers, but nothing works. Also, when a=1,b=7,c=8, we have f(1,7,8)<23=f(1,1,1). Hence f(a,b,c)≥2a3 for all a,b,c>0 is clearly not true. I have not tried f(a,b,c)≥2c3, but even if it is true, then you have to prove that RHS≤2c3 is order your solution to work.
The problem is that you are doing the following: Say you have correctly proven that for each x∈S, f(x)≥g(x). Then, you are claiming that f(x)≥max(g(x)x∈S), which is generally not true.