The smallest integer greater than $1$ is $2$, so we see that
$$\frac{b^3}{a^4} \ge 2 \quad \text{and} \quad \frac{a^3}{b^2} \ge 2.$$
It follows that
$$a = \left(\frac{b^3}{a^4}\right)^2 \cdot \left(\frac{a^3}{b^2}\right)^3 \ge 2^5 \quad \text{and} \quad b = \left(\frac{b^3}{a^4}\right)^3 \cdot \left(\frac{a^3}{b^2}\right)^4 \ge 2^7.$$
So we see that $a + b \ge 2^5 + 2^7 = 160$. On the other hand, we see that
$(a, b) = (2^5, 2^7)$ is a solution. So $160$ is the smallest possible value for $a + b$.