Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.

${}^{1}$Note that the measure of the integra $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$

Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.

${}^{1}$Note that, properly, we are defining the line integral in terms of three unit vectors normal to the line, so, the integral would have the form $\int \sqrt{|g|}\epsilon_{abcd}{\hat t^{a}}{\hat \theta^{b}}{\hat \phi^{c}}F^{d}$. When this is completely simplified, you'll find that the measure of the integral $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$, producing the term seen above.

Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}$. We should then note that this factor will exactly cancel the effect you cite.

Your naïve interpretation of the work equation doesn't quite make sense in this context. Consider that the standard formula for work gives that $W=\int {\vec F}\cdot d{\vec x}$. At minimum, the presence of the dot product in the above equation should not be ignored, and we should interpret, for a radial force, the work to be equal to $\int \frac{F\,dr}{\sqrt{1-\frac{2M}{r}}}^{1}$. We should then note that this factor will exactly cancel the effect you cite.

${}^{1}$Note that the measure of the integra $\sqrt{|g|}$ has a factor of $\sqrt{1-\frac{2M}{r}}$ that cancels against the factor of $\frac{1}{1-\frac{2M}{r}}$ in $g_{rr}$