First, we need to find the elements in the relation $R$ defined by $4x \le 5y$ where $x, y \in A = \{1, 2, 3, 4, 5\}$.

We can list the pairs $(x, y)$ that satisfy the condition:

• If $x = 1$, then $4(1) \le 5y \Rightarrow 4 \le 5y \Rightarrow y \ge \frac{4}{5}$. So, $y$ can be $1, 2, 3, 4, 5$.
• If $x = 2$, then $4(2) \le 5y \Rightarrow 8 \le 5y \Rightarrow y \ge \frac{8}{5}$. So, $y$ can be $2, 3, 4, 5$.
• If $x = 3$, then $4(3) \le 5y \Rightarrow 12 \le 5y \Rightarrow y \ge \frac{12}{5}$. So, $y$ can be $3, 4, 5$.
• If $x = 4$, then $4(4) \le 5y \Rightarrow 16 \le 5y \Rightarrow y \ge \frac{16}{5}$. So, $y$ can be $4, 5$.
• If $x = 5$, then $4(5) \le 5y \Rightarrow 20 \le 5y \Rightarrow y \ge \frac{20}{5}$. So, $y$ can be $4, 5$.

Therefore, $R = \{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)\}$.

The number of elements in $R$ is $m = 16$.

Now, we need to find the minimum number of elements to add to $R$ to make it symmetric. A relation is symmetric if whenever $(x, y) \in R$, then $(y, x) \in R$.

We have the following pairs in $R$:

• $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$ are already symmetric.
• $(1, 2) \in R$, we need $(2, 1)$.
• $(1, 3) \in R$, we need $(3, 1)$.
• $(1, 4) \in R$, we need $(4, 1)$.
• $(1, 5) \in R$, we need $(5, 1)$.
• $(2, 3) \in R$, we need $(3, 2)$.
• $(2, 4) \in R$, we need $(4, 2)$.
• $(2, 5) \in R$, we need $(5, 2)$.
• $(3, 4) \in R$, we need $(4, 3)$.
• $(3, 5) \in R$, we need $(5, 3)$.
• $(4, 5) \in R$, we need $(5, 4)$, but $(5,4)$ is already in R.

So, we need to add the following pairs: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$.

The number of elements to add is $n = 9$.

Finally, $m + n = 16 + 9 = 25$.