$g(n-1) \geq g(n) \Leftrightarrow (n+1) \sum f( \frac{i}{n-1}) \leq n \sum f( \frac{i}{n})$.
This follows by summing up the following inequlities:
$\begin{array} { l l l l l l l }
& & n \times f(0) & = & n \times f(0) &=& n \times f(0), \\
1 \times f(\frac{0}{n-1}) &+& (n-1) \times f( \frac{1}{n-1} ) & \geq & n \times f( \frac{\frac{0}{n-1} + \frac{n-1}{n-1}}{n}) &=& n f(\frac{1}{n}), \\
2 \times f( \frac{1}{n-1}) &+& (n-2) \times f( \frac{2}{n-1}) &\geq& n \times f ( \frac{\frac{ 2}{n-1} + \frac{2n-4}{n-1}}{n}) &=& n f(\frac{2}{n}), \\
\vdots \\
i \times f( \frac{i-1}{n-1}) & + & (n-i) \times f( \frac{i}{n-1}) & \geq & n \times f( \frac{\frac{ i^2 - i}{n-1} + \frac{ni - i^2}{n-1} } { n } ) & = & n f(\frac{i}{n}), \\
(n-1) \times f( \frac{n-2}{n-1} ) &+& 1 \times f( \frac{n-1}{n-1}) &\geq& n \times f(\frac{ \frac{ n^2-3n+2}{n-1} + \frac{n-1}{n-1} }{n}) &=& n f(\frac{n-1}{n}) \\
n \times f(1) & & &=& n \times f(1)&=& n \times f(1) \\
\end{array}$
Note: Strict convexity gives strict inequalities, so the function is strictly decreasing.