If , where and are positive integers and the maximum possible value of is , where and are relatively prime positive integers, then find .
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We can take S = a + b and rewrite a + b a b + 1 = S a ( S − a ) + 1 . Also, we assume a ≤ b , because of the symmetry in the expressions.
One can find out that if S > S ′ , then S a ( S − a ) + 1 > S ′ a ( S ′ − a ) + 1 , for all 1 ≤ a ≤ ⌊ 2 S ⌋ . It means if we plot S a ( S − a ) + 1 , with a as the free variable and S , S ′ fixed, then the plot of S a ( S − a ) + 1 is always above S ′ a ( S ′ − a ) + 1 and they only meet at a = 1 .
As we want to maximise a 3 + b 3 a 3 b 3 + 1 we may intuitively start with values of a , b that are closer to each other. Actually one can show that, fixing S , the max of S a ( S − a ) + 1 is achieved at a = ⌊ 2 S ⌋ . If one tries a = 2 , b = 3 then a + b a b + 1 = 5 7 < 2 3 . Then we can increase S to force S a ( S − a ) + 1 to get closer to 2 3 . If a = 3 , b = 3 then the inequality would not hold true. For a = 2 , b = 4 , a + b a b + 1 = 6 9 = 2 3 , which is not valid. Therefore, if a = 2 and S > 5 , according to the previous paragraph a + b a b + 1 ≥ 2 3 . If a = 1 then a + b a b + 1 = 1 , which is not the best we have found so far. We cannot have a > 2 , with S > 5 , because for a = 2 , b = 4 the fraction is already equal to 2 3 and for higher a the fraction would become even larger. So, a = 2 , b = 3 is the best we found and by substituting we get a 3 + b 3 a 3 b 3 + 1 = 5 3 1 and 3 1 + 5 = 3 6