Romanian Mathematical Olympiad

a) Since $f^{\prime }$ has the intermediate value property, it follows that $f^{\prime }(x)<1$, for every $x\in [0,1]$, or $f^{\prime }(x)>1$, for every $x\in [0,1]$. But, in the second case $f$ would be strictly increasing, which contradicts $f(0)=f(1)$. So $f^{\prime }(x)<1$, for every $x\in [0,1]$, whence $g:[0,1]\to \mathbb{R}$, $g(x)=f(x)-x$ has $g^{\prime }(x)=f^{\prime }(x)-1<0$, for all $x\in [0,1]$, that is $g$ is strictly decreasing.

b) Notice that $$s_{\Delta }(g)<{\int }_0^{1}g(x)dx<S_{\Delta }(g),$$ where $s_{\Delta }(g)$ and $S_{\Delta }(g)$ are the inferior, respectively superior Darboux sums of $g$ for the partition $\Delta =\{0,\frac{1}{n},\dots ,\frac{n-1}{n},1\}$. Then $$\frac{1}{n}\sum _{k=0}^{n-1}\left(f\left(\frac{k+1}{n}\right)-\frac{k+1}{n}\right)<{\int }_0^1(f(x)-x)dx<\frac{1}{n}\sum _{k=0}^{n-1}\left(f\left(\frac{k}{n}\right)-\frac{k}{n}\right).$$ Since $f(0)=f(1)$ and ${\int }_0^{1}f(x)dx=0$, this yields $$\sum _{k=0}^{n-1}f\left(\frac{k}{n}\right)-\frac{n+1}{2}<-\frac{n}{2}<\sum _{k=0}^{n-1}f\left(\frac{k}{n}\right)-\frac{n-1}{2},$$ whence the required inequality.

Alternative Solution (b): Let $G:[0,1]\to \mathbb{R}$, $G(x)=F(x)-\frac{1}{2}{x}^2$, where $F$ is a primitive of $f$. The function $G^{\prime }(x)=f(x)-x$ is strictly decreasing, because $G^{\prime \prime }(x)=f^{\prime }(x)-1<0$, for every $x\in [0,1]$. Now, Lagrange's mean value theorem, used for the function $G$ on the intervals $[\frac{k}{n},\frac{k+1}{n}]$, $k=0,\dots ,n-1$, yields $$f\left(\frac{k+1}{n}\right)-\frac{k+1}{n}<n\left(F\left(\frac{k+1}{n}\right)-F\left(\frac{k}{n}\right)-\frac{2k+1}{2n^2}\right)<f\left(\frac{k}{n}\right)-\frac{k}{n}.$$ These inequalities add up to $$\sum _{k=0}^{n-1}f\left(\frac{k+1}{n}\right)-\frac{n+1}{2}<n\left(F(1)-F(0)-\frac{1}{2}\right)<\sum _{k=0}^{n-1}f\left(\frac{k}{n}\right)-\frac{n-1}{2}.$$ Since $F(1)-F(0)={\int }_0^{1}f(x)dx=0$ and $f(0)=f(1)$, the required inequality follows immediately.