Problem: $f$ is a strictly monotonic and continuous function on $[0, 1]$, such that $f(0)=0$ and $f(1)=1$.
Then prove that $f(\frac{1}{10})+f(\frac{2}{10})+\cdots f(\frac{9}{10})+f^{-1}(\frac{1}{10})+f^{-1}(\frac{2}{10})+\cdots f^{-1}(\frac{9}{10}) \leq \frac{99}{10}$.
I am able show (clear if you draw a rough graph) that the LHS is less than 10. But I am unable to improve on it. I even did the entire inverse function chapter from Michael Spivak calculus for this problem. Any ideas and even hints would be much appreciated.
In fact the sum is strictly less than $99/10$, but it can be arbitrarily close to $99/10$, which is to say $99/10$ is indeed a strict upper bound.
Since $f$ is strictly increasing, $$\int_{j/10}^{(j+1)/10}f(t)\,dt>\frac{f(j/10)}{10}, $$so $$\sum_{j=1}^9f\left(\frac j{10}\right)<10\int_{1/10}^1f(t)\,dt.$$For exactly the same reason, $$\sum_{j=1}^9f^{-1}\left(\frac j{10}\right)<10\int_{1/10}^1f^{-1}(t)\,dt.$$But $\int_{1/10}^1f$ is the area of $$A_1=\{(x,y):1/10\le x\le 1,0<y<f(x)\},$$while if you tilt your head at the right angle you see that $\int_{1/10}^1f^{-1}$ is the area of $$A_2=\{(x,y):1/10\le y\le 1,0<x<f^{-1}(y)\},$$ Again since $f$ is increasing, $y<f(x)$ implies that $f^{-1}(y)<x$; hence $A_1$ and $A_2$ are disjoint. So $$\sum_{j=1}^9f\left(\frac j{10}\right)+\sum_{j=1}^9f^{-1}\left(\frac j{10}\right)\le 10\, area\,(A_1\cup A_2).$$But $$A_1\cup A_2\subset[0,1]\times[0,1]$$and $$(A_1\cup A_2)\cap([0,1/10)\times[0,1/10))=\emptyset,$$hence $$area\,(A_1\cup A_2)\le1-1/100.$$
So the sum is less than $99/10$. For an example showing this is best possible, let $\epsilon>0$ be very small. There is a strictly increasing continuous function $f$ with $f(0)=0$, $f(1)=1$, and such that for every integer $j$ with $1\le j\le 9$ we have $f(j/10-\epsilon)=j/10$ and $f(j/10)=(j+1)/10-\epsilon$. (Draw a picture connecting the dots with straight lines and you get a strictly increasing continuous function.) Hence $f^{-1}(j/10)=j/10-\epsilon$, and if you add it all up you see the sum is $99/10 - 18\epsilon$.