Given that are positive reals and .Find the minimum of the expression above.
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Relevant wiki: Cauchy-Schwarz Inequality
I am not sure if what I did is completely fine. Denote a + b + c = 3 u , a b + b c + c a = 3 v 2 , a b c = w 3 . We have u = t v 2 , t ≥ 1 . Now the expression we try to minimize is:
3 v 2 9 u 2 − 6 v 2 + 3 u = 3 v 2 9 t 2 v 4 − 6 v 2 + 3 t v 2 = 3 v 2 t 2 + t − 2 . Now since 3 v 2 > 0 we have in the RHS an increasing function of the variable t, which gets a minimum when t is minimal, thus, t = 1 or a + b + c = a b + b c + c a . In this point, the result of the expression will be 3 v 2 − 1 = 3 u − 1 . Thus, we want to minimize v 2 = u . Now we look on the constraint. It can be re-written as:
3 w + w 3 ≤ 3 u + w 3 ≤ 3 v 2 + w 3 ≤ 4 , or 3 w + w 3 ≤ 4 . Now since we want to maximize w in order to minimze u , we have 3 w + w 3 = 4 . The one real solution is w = 1 , and since the equality happens when a = b = c we have a = b = c = 1 , substituting will give us the result, 2.