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positive reals a,b,c give
ab + bc + ac = 3
Prove that:
a³ + b³ + c³ + 6abc ≥ 9
Can someone help??
ab + bc + ac = c(a+b) + ab = 3
c^3(a+b) + abc^2 = 3c^2
c^3 + c^2*ab/(a+b) = 3c^2/(a+b)
c^2(ab - 3)/(a+b) = c^3
(ab-3)/(a+b) = c
a^3 + b^3 + (ab-3)^3/(a+b)^3 + 6ab(ab-3)/(a+b) is supposedly >= 9
But since a^3 + b^3 and is the lowest when a=b and (ab-3)^3/(a+b)^3 + 6ab(ab-3)/(a+b) is negative for all positive numbers a,b,c hence the whole expresion is possibly minimum when a=b since ab is then max but < 3 since ab + bc + ac = 3 and since -ab/(a+b) is the largest negative when a and b are equal, we get:
2b^3 + (b^2-3)^3/(2b)^3 + 6b^2(b^2-3)/2b >= 9
b^6 + 1/16(b^2-3)^3 + 3b^5(b^2-3) <= 9
But since we also get that since b can be exchanged in this solution to c, c=a=b, and since they are equal when the equation yield minimum, and the only such solution when ab + bc + ac = 3 is when they are all one, we get that minimum is:
2b^3 + b³ + 6b^3 = (2 + 1 + 6)*1^3 = 9
I see clearly now, the universe have the black dots, Thus I am on my way of inventing this remedy...
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