$\sum (f(\frac{k}{n})) + \sum (f^{-1}(\frac{k}{n}))$ I would appreciate some help with the following Putnam preparatory problem (you can google Putnam 2010, Week 3, inequalities, problem 13):
Let $f$ be a continuous and monotonically increasing function such that $f(0) = 0$ and $f(1) = 1$. Prove that:
$f(0.1)+f(0.2)+···+f(0.9)+f^{−1}(0.1)+f^{−1}(0.2)+···+f^{−1}(0.9) ≤ 9.9$ (here $f^{−1}$ is the inverse function of $f$).
I've made two attempts and came up with either a more relaxed bound (10.0) or a tighter bound (9.5) but for a special case (when $f(x)$ doesn't change convexity). I can present those if anyone interested (I used $\int_a^b f(x)dx+\int_{f(a)}^{f(b)} f^{−1}(x)dx=bf(b)−af(a)$ and lower Darboux sums or trapezoidal method), but bottom line is I couldn't prove the general 9.9 bound.
Thanks in advance.
 A: I present and idea on how you could start: We start the same way as you probably already did, but then take care of a detail.
The core idea is following:
What we're trying to do here is making use of the fact that the red square is always ignored in the summation: (The yellow rectangles represent the sums.)


Surely $$\int_0^1 f(x) dx + \int_0^1 f^{-1}(x) dx = 1 \qquad (1)$$
Let $x_k = 0.1\cdot k$. Then define the piecewise constant step function $g(x) = f(x_k)$ for $ x \in [x_k,x_{k+1})$ and similarly $g^{-1}(x) = f^{-1}(x_k)$ for $x \in [x_k, x_{k+1})$. (Note that I just use the symbol $g^{-1}$ for symmetry reasons, but $g^{-1}$ does not denote the inverse of $g$.)Then surely $g(x) \leq f(x)$ and $g^{-1}(x) \leq f(x) \forall x \in [0,1]$.
Therefore 
$$\frac{1}{10}(f(0.1)+f(0.2)+\ldots+f(0.9))=\int_{0.1}^1 g(x) dx \leq \int_{0.1}^1 f(x)dx \qquad (2)$$
and 
$$\frac{1}{10}(f^{-1}(0.1)+f^{-1}(0.2)+\ldots+f^{-1}(0.9))=\int_{0.1}^1 g^{-1}(x) dx \leq \int_{0.1}^1 f^{-1}(x)dx \qquad (3)$$
Now note that $$G:=\int_0^{0.1} g(x) dx + \int_0^{0.1} g^{-1}(x)dx = 0 \qquad (4)$$
Now we just have to consider $$F:=\int_0^{0.1} f(x) dx + \int_0^{0.1} f^{-1}(x)dx \qquad (5)$$
So now we can add $(2)$ and $(3)$ and multiply by $10$ and get
$$\sum_{k=1}^9 f(x_k) + f^{-1}(x_k)  \overset{(2) \& (3)}{\leq} 10\int_{0.1}^1 f(x) dx + 10\int_{0.1}^1 f^{-1}(x) dx \overset{(1)\& (5)}{=} 10(1-F) $$
It remains to show that $F \geq 0.01$:
Let $a=0$ and $b=0.1$. 
Then $$\begin{align*}F  &=  \int_0^b f(x)dx+\int_{0}^{b} f^{−1}(x)dx  \\ 
 & \geq \int_0^b \min\{f(x),b\} dx+\int_{0}^{b} \min\{f^{−1}(x),b\}dx \\
& =b^2 = 0.01
\end{align*}$$
This completes the proof.
