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For convex figure $\Sigma$ in $\mathbb S^2$, the isoperimetrical inequality should look like $$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$

If $\Sigma$ and $\Sigma'$ are the intersections of $\mathbb S^2$ with your cones then by Crofton formula we get $$\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}+\frac{\mathop{\rm area}\Sigma'}{2\cdot\pi}=1$$ Hence te result.

P.S. The extreme values should be for round cone and positive octant, but I do not see a proof in higher dimensions.

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For convex figure $\Sigma$ in $\mathbb S^2$, the isoperimetrical inequality should look like $$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$

If $\Sigma=\Omega\cap\mathbb S^2$ \Sigma$and$\Sigma'=\Omega'\cap\mathbb \Sigma'$are the intersections of$\mathbb S^2$with your cones then $$\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}+\frac{\mathop{\rm area}\Sigma'}{2\cdot\pi}=1$$ Hence te result. P.S. The extreme values should be for round cone and positive octant, but I do not see a proof in higher dimensions. 1 For convex figure$\Sigma$in$\mathbb S^2$, the isoperimetrical inequality should look like $$\left(\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}\right)^2+\left(1-\frac{\mathop{\rm area}\Sigma}{2\cdot\pi}\right)^2\ge 1.$$ If$\Sigma=\Omega\cap\mathbb S^2$and$\Sigma'=\Omega'\cap\mathbb S^2\$ then $$\frac{\mathop{\rm perim}\Sigma}{2\cdot\pi}+\frac{\mathop{\rm area}\Sigma'}{2\cdot\pi}=1$$ Hence te result.