Determine the least real number M such that the inequalityholds for all real numbers a, b, and c.

gee... which company uses IMO problem for interview?this is from IMO 2006 problem 3. the answer is M=9sqrt(2)/32.the tricky part is that equality holds not when any two of a, b, c are equal (i doubt anyone can guess the answer in ten minutes...) but if you set x=a-b, y=b-c, z=c-a, s=a+b+c, you could turn the original asymmetric expression into symmetric, and standard techniques follow. anyone interested could just google "IMO 2006"

i'm impressed! i hope you didn't google the technique and answer QuoteOriginally posted by: wileyswgee... which company uses IMO problem for interview?this is from IMO 2006 problem 3. the answer is M=9sqrt(2)/32.the tricky part is that equality holds not when any two of a, b, c are equal (i doubt anyone can guess the answer in ten minutes...) but if you set x=a-b, y=b-c, z=c-a, s=a+b+c, you could turn the original asymmetric expression into symmetric, and standard techniques follow. anyone interested could just google "IMO 2006"

i just think the question is not very well suited for an interview. it's the last problem of the first day, which is supposed to be very hard. besides knowing calculus here does not really help much (Lagrangian multiplier works, but much much more tedious than using AM-GM).but i don't mind ppl thinking about it independently (that's why i just gave M, but not the proof - it's still pretty hard to prove the inequality even you know M).

i totally agree...it was for an interview at a hedge fund. needless to say, i didn't get the answer, but i did get the offer...i guess they were just interested in seeing how people think and didn't expect anyone to really get it, unless you're one of the few geniuses.QuoteOriginally posted by: wileyswi just think the question is not very well suited for an interview. it's the last problem of the first day, which is supposed to be very hard. besides knowing calculus here does not really help much (Lagrangian multiplier works, but much much more tedious than using AM-GM).but i don't mind ppl thinking about it independently (that's why i just gave M, but not the proof - it's still pretty hard to prove the inequality even you know M).

thx for clarifying, and congrats. i think even you could have come up with the answer, they would have suspected that you knew the origin of the problem...

QuoteOriginally posted by: thisfrenchkissDetermine the least real number M such that the inequalityholds for all real numbers a, b, and c.An observation:We can normalize the variables as follows:Set r=sqrt((a^2)+(b^2)+(c^2)).Set A=a/r, so a=rA. Similarly we have b=rB, and c=rC.Then the inequality becomes:| AB((A^2)-(B^2)) + BC((B^2)-(C^2)) + CA((C^2)-(A^2)) | <= M, plus the condition (A^2) + (B^2) + (C^2) = 1.

yes the lagrange multiplier works, but lots of algebra... another way might be spherical coordinates? but it's not clear how that helps.

Here is one simple, but less tight lower bound:(a^2 + b^2 + c^2 )^2 - (1/M)(ab(a^2-b^2)...) = a^4 + b^4 + 2a^2b^2 - (1/M)ab^3 +(1/M)a^3b ... fill in similar terms containing b, c and c,aNote that:a^4 + b^4 + c^4 + 2a^2b^2 - Mab^3 +Ma^3b is of the form (a^2 + b^2 + k * a * b)^2, if we can find the right kso, we complete squares as follows: (a^2 /sqrt(2) - b^2/sqrt(2) +sqrt(3) * a*b)^2 = a^4/2 + b^4/2 +2a^2b^2 - sqrt(3*2)(ab^3 - a^3b) + symmetric termsThis is clearly greater than 0, therefore, for any M = 1/sqrt(6), the inequality works. For M > 1/sqrt(6), the inequality should certainly work.1/sqrt(6) works out to be .408, as opposed to 9/(16 (sqrt(2))) = 0.397 !

thanks. this might be the best you could do for an interview.

Terrible!