Let $f(x)$ be bounded on $[a,b]$ . Let $P_0$ be a partition such that: $P = \{a=t_0,t_1,\dots t_n=b\}$, where $a=t_o<t_1<t_2\dots t_n=b$ and $\;t_k= a+ \frac{k(b-a)}{n}$, for all $k$. Intuitively this $P_0$ seems to be the partition that gives the lowest value (for a fixed $n$) of $U(f,P)$ (the upper sum). I am having trouble proving this.
I tried to proceed as follows: Consider a new partition $P'$generated by disturbing $P_0$ at one particular point $t_k$ such that the the new partition is the same except $t_k$ replaced by $t_k + \Delta t$. Since any arbitrary partition containing $n$ intervals can be generated by disturbing $P_0$ at finitely many points, it suffices to show that $U(f,P_0) \leq U(f,P')$.
I am unable to prove the required result. Any suggestions to prove or disprove the statement are much appreciated.