A number theory problem by Prem Chebrolu

If a b + 1 a + b < 3 2 \dfrac{ab + 1}{a + b}< \dfrac{3}{2} , where a a and b b are positive integers and the maximum possible value of a 3 b 3 + 1 a 3 + b 3 \dfrac{a^{3}b^{3} + 1}{a^{3} + b^{3}} is x y \dfrac{x}{y} , where x x and y y are relatively prime positive integers, then find x + y x + y .


The answer is 36.

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2 solutions

We can take S = a + b S=a+b and rewrite a b + 1 a + b = a ( S a ) + 1 S \frac{ab+1}{a+b} = \frac{a(S-a)+1}{S} . Also, we assume a b a \leq b , because of the symmetry in the expressions.

One can find out that if S > S S>S' , then a ( S a ) + 1 S > a ( S a ) + 1 S \frac{a(S-a)+1}{S} > \frac{a(S'-a)+1}{S'} , for all 1 a S 2 1 \leq a \leq \lfloor \frac{S}{2}\rfloor . It means if we plot a ( S a ) + 1 S \frac{a(S-a)+1}{S} , with a a as the free variable and S , S S,S' fixed, then the plot of a ( S a ) + 1 S \frac{a(S-a)+1}{S} is always above a ( S a ) + 1 S \frac{a(S'-a)+1}{S'} and they only meet at a = 1 a=1 .

As we want to maximise a 3 b 3 + 1 a 3 + b 3 \frac{a^3b^3+1}{a^3+b^3} we may intuitively start with values of a , b a,b that are closer to each other. Actually one can show that, fixing S S , the max of a ( S a ) + 1 S \frac{a(S-a)+1}{S} is achieved at a = S 2 a= \lfloor \frac{S}{2} \rfloor . If one tries a = 2 , b = 3 a=2,b=3 then a b + 1 a + b = 7 5 < 3 2 \frac{ab+1}{a+b}= \frac{7}{5} < \frac{3}{2} . Then we can increase S S to force a ( S a ) + 1 S \frac{a(S-a)+1}{S} to get closer to 3 2 \frac{3}{2} . If a = 3 , b = 3 a=3,b=3 then the inequality would not hold true. For a = 2 , b = 4 a=2,b=4 , a b + 1 a + b = 9 6 = 3 2 \frac{ab+1}{a+b}= \frac{9}{6}=\frac{3}{2} , which is not valid. Therefore, if a = 2 a=2 and S > 5 S>5 , according to the previous paragraph a b + 1 a + b 3 2 \frac{ab+1}{a+b} \geq \frac{3}{2} . If a = 1 a=1 then a b + 1 a + b = 1 \frac{ab+1}{a+b}=1 , which is not the best we have found so far. We cannot have a > 2 a>2 , with S > 5 S>5 , because for a = 2 , b = 4 a=2,b=4 the fraction is already equal to 3 2 \frac{3}{2} and for higher a a the fraction would become even larger. So, a = 2 , b = 3 a=2,b=3 is the best we found and by substituting we get a 3 b 3 + 1 a 3 + b 3 = 31 5 \frac{a^3b^3+1}{a^3+b^3}=\frac{31}{5} and 31 + 5 = 36 31+5=36

Barr Shiv
Oct 11, 2018

First step is to simplify the inequailty till you get from one side of the inequailty an Integer and the other side a product, it's exactly if one would solve a DP eqaution. after some algebra we get: (2a-3)(2b-3)<5 that's quite brilliant becuase we can see from the expression that if we merely choose a>3 and b>3 this inequality breaks down. that's a pivotal step because that really narrows down the options for a and b. so there's a particular range of solutions that we want to find, first let's solve for (2a-3)(2b-3)=5 we know that 5 is a prime so the only possibilties for the factors are 1 and 5 and since the Symmetry of the equation we can choose which 1 possibility and then we know that the other solution is just switching the values of a and b. so solving for this equation we get: a=2/4 b=4/2. we can find only two more posibillties in which the product is positive and these:(2,2) and (2,3) for a or b.

very important to notice that if: a=1 then for every value of b>1 (and vise versa) the inequailty holds. but pluging them in the expression we get: (b^3+1)/(b^3+1)=1 and vise versa for a. which leaves us with only checking (2,2) and (3,3) or just 1 and see what is bigger and that's the maximal value of the expression. after pluging in we see that:1< F(2,2)<F(3,2) which means that F(3,2) is indeed the max value which is 62/10=31/5 x+y=31+5=36

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