If $f$ is a continuous function on $[0,1]$ such that for all $t_1, t_2 \in [0,1]$,
$$f\left(\frac{t_1+t_2}{2}\right) \leq \frac{f(t_1)+f(t_2)}{2}$$
Show that $$f\left(\frac{t_1+t_2+ \cdots +t_n}{n}\right)\leq \frac{f(t_1)+f(t_2)+\cdots f(t_n)}{n}$$
I tried proving this statement by induction. I could do this for even $n$.
Let $k=n/2$, then, \begin{align} f\left(\frac{t_1+t_2+ \cdots +t_n}{n}\right) &= f\left(\frac{t_1 + \cdots +t_k}{n} + \frac{t_{k+1} + \cdots +t_n}{n}\right) \\ &=f\left( \frac{\frac{2}{n}(t_1 + \cdots +t_k)}{2}+\frac{\frac{2}{n}(t_{k+1} + \cdots +t_n)}{2}\right)\\ &\leq \frac{f\left(\frac{2}{n}(t_1 + \cdots +t_k)\right)+f\left(\frac{2}{n}(t_{k+1} + \cdots +t_n)\right)}{2}\\ &=\frac{f(\frac{t_1+\cdots +t_k}{k})+f(\frac{t_{k+1}+\cdots +t_n}{k})}{2}\\ &\leq\frac{f(t_1)+f(t_2)+\cdots f(t_n)}{2k}\\ &=\frac{f(t_1)+f(t_2)+\cdots f(t_n)}{n} \end{align}
However, I don't know how to split should $n$ be odd. One reasonable thing I thought of was to split the middle term. For instance, if $n=3$, We could split it as such $t_1 + 0.5 t_2$ and $0.5 t_2 + t_3$. However, I couldn't find a neat way to carry out the induction if I do it like this. Can someone give a hint?