A number theory problem by Prem Chebrolu

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

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

As we want to maximise $\frac{a^3b^3+1}{a^3+b^3}$ 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 $\frac{a(S-a)+1}{S}$ is achieved at $a= \lfloor \frac{S}{2} \rfloor$ . If one tries $a=2,b=3$ then $\frac{ab+1}{a+b}= \frac{7}{5} < \frac{3}{2}$ . Then we can increase $S$ to force $\frac{a(S-a)+1}{S}$ to get closer to $\frac{3}{2}$ . If $a=3,b=3$ then the inequality would not hold true. For $a=2,b=4$ , $\frac{ab+1}{a+b}= \frac{9}{6}=\frac{3}{2}$ , which is not valid. Therefore, if $a=2$ and $S>5$ , according to the previous paragraph $\frac{ab+1}{a+b} \geq \frac{3}{2}$ . If $a=1$ then $\frac{ab+1}{a+b}=1$ , 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 $\frac{3}{2}$ 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 $\frac{a^3b^3+1}{a^3+b^3}=\frac{31}{5}$ and $31+5=36$

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