Consider a Riemannian sphere $(S^2,g)$. The Cheeger constant of $(S^2,g)$ is $$ h=\inf_{\gamma} \frac{L(\gamma)}{\min\{S(A_1),S(A_2)\}} $$ where $\gamma$ is closed curve on $S^2$ which divide $S^2$ into two disjiont parts $A_1$ and $A_2$. $L(\gamma)$ is the length of $\gamma$. $S(A_i)$ is the area of $A_i$ (i=1,2). The Cheeger set is the curve $\gamma$ attaining the $\inf$, namely $$ \frac{L(\gamma)}{\min\{S(A_1),S(A_2)\}}=h $$
Assuming $\gamma$ is the Cheeger set of $(S^2,g)$, and $S(A_1)=S(A_2)$. Then, I feel that $\gamma$ will attain $$ \inf_\gamma L(\gamma)\left(\frac{1}{S(A_1)} + \frac{1}{S(A_2)}\right) $$ where $\gamma$ is closed curve on $S^2$ which divide $S^2$ into two disjiont parts $A_1$ and $A_2$. But I don't know how to show it.