3 added 78 characters in body

I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 4, 487–507 and Yurii D. Burago and Victor A. Zalgaller, "Geometric inequalities".

I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by $$dy = s(u,r) \,du \,dr$$ when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of Günther's inequality).

$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then
$$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$ and $$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$

Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.

I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in a an arXiv paper with Greg Kuperbergthat shall appear tomorrow on the arXiv, "A refinement of Günther's candle inequality" [arXiv:1204.3943].

2 Corrected the Burago-Zalgaller

I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 4, 487–507 and Dmitri Yurii D. Burago and Victor A. Zalgaller, "Geometric inequalities".

I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by $$dy = s(u,r) \,du \,dr$$ when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of Günther's inequality).

$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then
$$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$ and $$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$

Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.

I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in a paper with Greg Kuperberg that shall appear tomorrow on the arXiv.

1

I do not know if there is a way to get the isoperimetric inequality from the spectral gap, but both can be proven in almost the same way. The classical references for the linear isoperimetric inequality are S.-T. Yau, "Isoperimetric constants and the first eigenvalue of a compact Riemannian manifold", Ann. Sci. École Norm. Sup. (4) 8 (1975), no. 4, 487–507 and Dmitri Burago and Victor A. Zalgaller, "Geometric inequalities".

I like this proof so let me give it here (this is Burago-Zalgaller presentation). For any unit tangent vector $u$ and positive real $r$, let $s(u,r)$ be the "candle function" defined by $$dy = s(u,r) \,du \,dr$$ when $y=\exp_x(ru)$ and $u\in UT_xM$. Up to a normalization, this is simply the jacobian of the exponential map. The curvature hypothesis implies $(\log s(u,r))'\geqslant \sqrt{-\kappa}(n-1)$ where the prime denotes derivative with respect to $r$ (this is a consequence of Günther's inequality).

$\Omega$ is contained in the union of all geodesic rays from any fixed point $x_0$ to $\partial \Omega$. Let $U\subset UT_{x_0}M$ be the set of unit vectors generating geodesics that intersect $\Omega$, and for $u\in U$ let $r_u$ be the last intersection time of the geodesic generated by $u$ with $\Omega$. Then
$$\mathrm{Vol}(\partial \Omega) \geqslant \int_U s(u,r_u) \,du$$ and $$\mathrm{Vol}(\Omega) \leqslant \int_U \int_0^{r_u} s(u,t) \,dt\,du.$$

Now, writing $s(u,r_u)=\int_0^{r_u} s'(u,t) \,dt$ and using Günther's inequality, the desired result comes.

I cannot help self-advertising: in fact, the same conclusions (Günther inequality, hence both the linear isoperimetric inequality and the spectral gap of MCKean) hold under a weaker curvature bound (some higher but non-positive sectional curvature can be compensated by enough more negative sectional curvature in other directions). This is explained in a paper with Greg Kuperberg that shall appear tomorrow on the arXiv.