Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum. We fix $m$ in $\mathbb{N}^{*}$ and $\varepsilon$ in $\mathbb{R}^{+*}$. Using the Euclidean division of $n$ by $m$, show that there exists an integer $N$ such that for all $n > N$, $$\frac{u_{n}}{n} \geqslant \frac{u_{m}}{m} - \varepsilon$$
Let $\left(u_{n}\right)_{n \geqslant 0}$ be a real sequence such that: $\forall (m,n) \in \mathbb{N}^{2}, \quad u_{m+n} \geqslant u_{m} + u_{n}$. We suppose that the set $\left\{\frac{u_{n}}{n}, n \in \mathbb{N}^{*}\right\}$ is bounded above and we denote by $s$ its supremum.
We fix $m$ in $\mathbb{N}^{*}$ and $\varepsilon$ in $\mathbb{R}^{+*}$. Using the Euclidean division of $n$ by $m$, show that there exists an integer $N$ such that for all $n > N$,
$$\frac{u_{n}}{n} \geqslant \frac{u_{m}}{m} - \varepsilon$$