
A 2. Find the maximum positive integer $k$ such that for any positive integers $m, n$ such that $m^{3}+n^{3}>$ $(m+n)^{2}$, we have

$$
m^{3}+n^{3} \geq(m+n)^{2}+k
$$


Solution. We see that for $m=3$ and $n=2$ we have $m^{3}+n^{3}>(m+n)^{2}$, thus

$$
3^{3}+2^{3} \geq(3+2)^{2}+k \Rightarrow k \leq 10
$$

We will show that $k=10$ is the desired maximum. In other words, we have to prove that

$$
m^{3}+n^{3} \geq(m+n)^{2}+10
$$

The last inequality is equivalent to

$$
(m+n)\left(m^{2}+n^{2}-m n-m-n\right) \geq 10
$$

If $m+n=2$ or $m+n=3$, then $(m, n)=(1,1),(1,2),(2,1)$ and we can check that none of them satisfies the condition $m^{3}+n^{3}>(m+n)^{2}$.

If $m+n=4$, then $(m, n)=(1,3),(2,2),(3,1)$. The pair $(m, n)=(2,2)$ doesn't satisfy the condition. The pairs $(m, n)=(1,3),(3,1)$ satisfy the condition and we can readily check that $m^{3}+n^{3} \geq(m+$ $n)^{2}+10$.

If $m+n \geq 5$ then we will show that

$$
m^{2}+n^{2}-m n-m-n \geq 2
$$

which is equivalent to

$$
(m-n)^{2}+(m-1)^{2}+(n-1)^{2} \geq 6
$$

If at least one of the numbers $m, n$ is greater or equal to 4 then $(m-1)^{2} \geq 9$ or $(n-1)^{2} \geq 9$ hence the desired result holds. As a result, it remains to check what happens if $m \leq 3$ and $n \leq 3$. Using the condition $m+n \geq 5$ we have that all such pairs are $(m, n)=(2,3),(3,2),(3,3)$.

All of them satisfy the condition and also the inequality $m^{2}+n^{2}-m n-m-n \geq 2$, thus we have the desired result.
