It's not obvious, but it follows by a straightforward argument using formula (1.1) in the Bulletin survey article by Lee and Parker. The case $n = 2$ is trivial, so assume $n > 2$. If a metric $\bar{g}$ is conformal to the metric $g$, then there exists a positive function $u$ such that $$ \bar{g} = u^{\frac{2}{n-2}}g $$ A straightforward calculation using (1.1) from Lee-Parker shows that $$ \int S(\bar{g})\,dV(\bar{g}) \ge \int uS(g)\,dV(g), $$ where $S(g)$ is the scalar curvature of $g$.substituting this into the definition of the Yamabe functional of $\bar{g}$ and applying Holder's inequality.

It's not obvious, but it follows by a straightforward argument using formula (1.1) in the Bulletin survey article by Lee and Parker. The case $n = 2$ is trivial, so assume $n > 2$. If a metric $\bar{g}$ is conformal to the metric $g$, then there exists a positive function $u$ such that $$ \bar{g} = u^{\frac{2}{n-2}}g $$ A straightforward calculation using (1.1) from Lee-Parker shows that $$ \int S(\bar{g})\,dV(\bar{g}) \ge \int uS(g)\,dV(g), $$ where $S(g)$ is the scalar curvature of $g$. The inequality now follows by substituting this into the definition of the Yamabe functional of $\bar{g}$ and applying Holder's inequality.

It's not obvious, but it follows by a straightforward argument. The case $n = 2$ is trivial, so assume $n > 2$. If a metric $\bar{g}$ is conformal to the metric $g$, then there exists a positive function $u$ such that $$ \bar{g} = u^{\frac{2}{n-2}}g $$ A straightforward calculation shows that $$ \int S(\bar{g})\,dV(\bar{g}) \ge \int uS(g)\,dV(g), $$ where $S(g)$ is the scalar curvature of $g$.substituting this into the definition of the Yamabe functional of $\bar{g}$ and applying Holder's inequality.

It's not obvious, but it follows by a straightforward argument using formula (1.1) in the Bulletin survey article by Lee and Parker. The case $n = 2$ is trivial, so assume $n > 2$. If a metric $\bar{g}$ is conformal to the metric $g$, then there exists a positive function $u$ such that $$ \bar{g} = u^{\frac{2}{n-2}}g $$ A straightforward calculation using (1.1) from Lee-Parker shows that $$ \int S(\bar{g})\,dV(\bar{g}) \ge \int uS(g)\,dV(g), $$ where $S(g)$ is the scalar curvature of $g$.substituting this into the definition of the Yamabe functional of $\bar{g}$ and applying Holder's inequality.