See the attachment
Do you have the answer? Is it 4/153 ?
No wait, 4/51 ?
I don't have the answer. Probably your solution would work
I just used symmetry , a=b=c
Sure that would work ?
not sure! :|
@RnR
Would that be required to prove
I dont know. :/
I was saying 0.784314 approx i.e. 4/51 , but guess the ans is 0.7082808
So my logic must be flawed ? @RnR
MAY BE
oh wait! my answer is correct !
it says all non negative numbers.
But, remember we will have to prove and justify each and every step.
hmm..I don't know the precise reasoning, but in most questions like these, symmetry works i.e. a=b=c. As I said, I don;t know the precise reasoning .
Well I accept what you said about symmetry; but without reasons that wouldn't work.
@ganeshie8
@mukushla can surely guide you, but he;s not online.. hmm @experimentX also!
well you may want to wait till the experts tune in..
this is not easy ... if it has numerical values, then Lagrange multiplies should work.
@DarthTony How can you say that a=b=c=x 3x=9/2, x=1.5 minimum = 3/2 First prove that this statement is true.
I think i saw similar Q on M.SE ... let me google it.
Whats that?
@experimentX whats M.SE
Ohhhhhhhhhh
I don't think you can just solve it by AM-GM type or Cauchy Swartz type
finding the minimum value and proving it are two different cases.
there's similar question here http://math.stackexchange.com/questions/191431/inequality-fracaa23-fracbb23-leq-frac12?rq=1
@dan815 I accept what you say in pts. 1 and 2 but still I don't find it sufficient to say that a=b=c will satisfy our answer. This is a difficult question and could not be solved with that ease.
here is a similar question ... this says to look for Cauchy-Swarz
can it be possible that a=b=0 and c=9/2.so that we get the minimum value??o is a non negative number
looks like this Q wen't unsolved in MSE too http://math.stackexchange.com/questions/275208/the-least-value-for-fracab354-fracbc354-fracca354
what class is this from?
@Peter14 No class
looks like Olympiad style Q.
if it's just min value then try making use of Lagrange Multiplies. @DarthTony the criteria is slight different a+b+c = 9/2 .... so this might be equal your answer. If it's just finding the value without caring method, then Lagrange multiplies also work.
@experimentX Correct said. Its an olympiad question
\[ f(a,b,c)=\frac{a}{b^3+54}+\frac{b}{c^3+54}+\frac{c}{a^3+54}\\ g(a,b,c)=a+b+c-{9\over2}\\ f_a=\lambda g_a=\lambda\\ f_b=\lambda g_b=\lambda\\ f_c=\lambda g_c=\lambda\\ a+b+c={9\over2} \] this is the set of langrangian equations the first three derivatives must be fairly simple (each giving a relation between two variables and \(\lambda\)).
like this: the first equ.\[{1\over b^3+54}-{3a^2c\over (a^3+4)^2}=\lambda\]
hmm.. no global minima!! lets see.. are the partial derivatives positive?
sorry again ... i was putting 24 against 54 http://www.wolframalpha.com/input/?i=Minimize%5B%7Ba%2F%28b%5E3+%2B+54%29+%2B+b%2F%28c%5E3+%2B+54%29+%2B+c%2F%28a%5E3+%2B+54%29%2C++++a+%2B+b+%2B+c+%3D%3D+9%2F2+%26%26+a+%3E%3D+0+%26%26+b+%3E%3D+0+%26%26+c+%3E%3D+0%7D%2C+%7Ba%2C+b%2C+c%7D%5D
if a function is bounded by numerical values, then Lagrange multiplies should work.
who is cheerful enough in the morning to continue the expansion? -> the real slim shady math guys say aye
i wonder if it can be proved by Cauchy Schwarz. If it's just asking for the value then Lagrange multiplies as well as guessing like symmetry argument. @electrokid i give up on lagrange
gotta go for now.
@SheldonEinstein is this a numerical problem or an analytical one?
Can't say as it is an olympiad one.
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