After reading the Wikipedia page on the spherical isoperimetric inequality, I came up with the following inequality by calculating some examples.
Consider a Riemannian metric $g$ on the topological sphere $S^2$. Let $L$ denote the length of any simple closed curves that divides $S^2$ into two parts with areas $A_1$ and $A_2$. The total area is $A=A_1+A_2$, and let $B=\min\{A_1,A_2\}$. Then, for any such curves, I guess $$ \frac{L^2}{B^2} \le \frac{4\pi}{B} -\frac{4\pi}{A} \tag{1} $$ when $\frac{L}{B}$ is at its minimum. I want to know how to prove it.