For continuous, monotonically-increasing $f$ with $f(0)=0$ and $f(1)=1$, prove $\sum_{k=1}^{10}f(k/10)+f^{-1}(k/10)\leq 99/10$

Question

A question from Leningrad Mathematical Olympiad 1991:

Let $f$ be continuous and monotonically increasing, with $f(0)=0$ and $f(1)=1$. Prove that:$$ \text{f}\left( \frac{1}{10} \right) +\text{f}\left( \frac{2}{10} \right) +...+\text{f}\left( \frac{9}{10} \right) +\text{f}^{-1}\left( \frac{1}{10} \right) +\text{f}^{-1}\left( \frac{2}{10} \right) +...+\text{f}^{-1}\left( \frac{9}{10} \right) \leqslant \frac{99}{10} $$

I tried to express them in areas to find inequalities but failed.

Answer 1

Just to convert @LeBlanc's comment into an answer:

For $1\le i\le 9$ let $S_i$ denote the rectangle with base $x\in[\tfrac{i}{10},\,\tfrac{i+1}{10})$ and height $f(\tfrac{i}{10})$, and let $P_i$ denote the rectangle with base $y\in[\tfrac{i}{10},\,\tfrac{i+1}{10})$ and height $f^{-1}(\tfrac{i}{10})$. These $18$ areas don't overlap; $S:=\bigcup_iS_i$ is a subset of the area under $y=f(x)$, while $P:=\bigcup_iP_i$ is a subset of the area in the square $[0,\,1]^2$ above $y=f(x)$. What's more, any point in any $P_i$ has $x\ge\frac{1}{10}$, while any point in any $S_i$ has $y\ge\frac{1}{10}$. Thus $$P\cup S\subseteq[0,\,1)^2\setminus[0,\,\tfrac{1}{10})^2.$$The desired sum is $10$ times the area of $P\cup S$, completing the proof.

LeBlanc's diagram includes the above exposition; hopefully the link will work indefinitely.

Answer 2

Note: In the title of the question the upper limit of $k$ needs to be 9.

When $y=f(x)$ is monotonically increasing in domain $D$ then area under the curves $y=f(x)$ and $y=f^{-1}(x)$ is more that those of the resprctive rectangles below them. Also the sum the area integrals $$A=\int_{x_1}^{x_2} f(x)~ dx+\int_{y_1}^{y_2} f^{-1} (y)~ dy = x_2 y_2 -x_1 y_1$$ Here $f: [0,1]\rightarrow[0,1]$. Let the domain be divided into 10 strips of width $1/10$. Let $S$ denotes area under the rectangles of equal width. Then $$S= \frac{1}{10}\sum_{k=1}^{9} [f(k/10)+ f^{-1}(k/10)] \le \sum_{k=1}^{9} \frac{(k+1)^2-(k)^2}{100}= \sum_{k=1}^{9}\frac{2k+1}{100}=\frac{99}{10}$$ $$\Rightarrow \sum_{k=1}^{10} [f(k/10)+ f^{-1}(k/10)] \le \frac{99}{10}. $$ As $S$ is the sum of area of rectangles below the curves $y=f(x)$ and $y=f^{-1}(x)$.