In about 1970s, Jaeger and Payan had proved the Nordhaus-Gaddum-type inequalities for domination numbers $\gamma(G)$ and $\gamma(\overline{G})$ of a graph $G$ and its complement $\overline{G}$ as:
For a graph $G$ of order $n \geq 1$,
$$ \min \{ 3, n+1 \} \leq \gamma(G) + \gamma(\overline{G}) \leq n+1,$$ and $$ \min \{2, n\} \leq \gamma(G) \cdot \gamma(\overline{G}) \leq n.$$
They stated that the upper bound of the product implies the upper bound of the sum.
I'm curious about this. I have tried using the fact that $(\gamma(G) + \gamma(\overline{G}))^2 \geq 0$, but it doesn't quite make it.
Would you mind provide me some hints to show that the upper bound of product implies the upper bound of the sum? Other than the fact that both bounds are sharp if $G = K_n$.
Or should I use the contraposition that $ \gamma(G) + \gamma(\overline{G}) > n+1$ implies $\gamma(G) \cdot \gamma(\overline{G}) > n$?
Consider the value $$\alpha := \max \gamma + \overline{\gamma} \\ \text{s.t} \quad \gamma \overline{\gamma} \leq n, \, \gamma \geq 1, \, \overline{\gamma} \geq 1, \, \gamma, \, \overline{\gamma} \in \mathbb{Z}$$ where we used that the domination number of a graph is an integer greater or equal to 1.
If we ignore the integer constraints on $\overline{\gamma}$, it follows that an optimal solution must have $\overline{\gamma} = n / \gamma$. Then $$\alpha \leq \max \gamma+\frac{n}{\gamma} \\ \quad \quad \text{s.t.} \quad 1 \leq \gamma \leq n, \, \gamma \in \mathbb{Z}$$ This maximization problem is an upper bound for $\alpha$ since we ignored $\overline{\gamma} \in \mathbb{Z}$. It is easy to see that this upper bound on $\alpha$ equals $n+1$ for all $n$. This shows that the upper bound on the sum is implied by the upper bound on the product.