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Algebra
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We have
1 = a3 + b3 + c3 = (a+b)3 - 3ab(a+b) +c3 = 1-3ab + c3 (since a+b = 1).
Therefore, we get 3ab = c3. Let S = a2 + b2 + c2. Then, S >= 0. Transform S as follows:
S = (a+b)2 -2ab + c2 = 1-2ab + c2 = 1 + c2 -(2/3)c3.
Thus, we get S as a function of c. To find the extremumu set S' = 0 which gives
S' = 2 c - 2 c2 = 0 => c = 0 or 1. It can be easily checked that the maximum is attained at c = 1 ( I have assumed that c is a positive number otherwise there won't be any maximum.) Hence, the maximum value of S becomes 4/3.
As kaymant sir pointed out, I have missed noting that the numbers are positive.
The answer is 
rajat came close, when he got the max value of c as 
. Only you have to also note that this maximum is obtained when 
I went about it in a similar manner as kaymant sir. Pls see http://www.mathlinks.ro/viewtopic.php?t=219459
So, the only thing to note here is that the function is increasing and the max is attained at the extremum which is not 1 but 1/4 if you choose ab as the variable and if you choose c
. Yes, indeed the maximum value will be (1/2) + (3/4)2/3. I missed the fact that c is bounded above, because c3 = 3 ab <= 3/4. And hence, c will never go up to 1.Preparing for IIT-JEE ?
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Applying cauchy:
(a^3 + b^3 + c^3)(a +b +c) >= (a^2+b^2+c^2)^2
=> root(1+c) >= (a^2+b^2 +c^2)
Now a^3 + b^3 = (a+b)(a^2 + b^2 + 2ab- 3ab)
c^3 = 1 - [ (a +b)^2 -3ab)
c^3 = 1- (1 -ab)
c^3 = 3ab
So max of c = 3/4
Hence max of wanted expression = root (7/4)