How to solve this problem involving the definite integral of a convex function?

Question

Let $i, j \in \mathbf{Z}, i<j$. Let $g:[i, j+1] \rightarrow \mathbf{R}$ be a decreasing convex function that satisfies $$ \begin{gathered} |g(z)-g(w)| \leq \frac{1}{2}|z-w|, \quad \forall z, w \end{gathered} $$ \begin{gathered} g(j+1) \leq 0 \leq g(j) \leq \cdots \leq g(i+1) \leq 1 . \end{gathered} $$ $$ For $a \in \mathbf{R}$, define $\lfloor a\rfloor:=\max (b \in \mathbf{Z}: b \leq a)$. Prove that $$ \frac{1}{2}\left\lfloor g(i)+\frac{1}{4}\right\rfloor+\sum_{m=i+1}^{j-1}\left\lfloor g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor g(j)+\frac{1}{4}\right\rfloor \leq \int_i^j g(z) d z . $$
My idea is： for simplicity, we can set $i=0, j=n$. $$ \text { let } g(k) \geqslant \frac{3}{4}, g(k+1)<\frac{3}{4}, k \in(0, n) \text {. } $$ The left side of the inequality is: $\frac{1}{2}+k$
Using the property of the convex function for the right side: $\frac{\int_0^n g(z) d z}{n} \geq \frac{g(k)+g(n)}{n-k}$
Thus, we want to prove: $n \cdot \frac{g(k)+g(n)}{n-k} \geq \frac{1}{2}+k$
But I don't know how to apply the other conditions, so I'm unsure about this.