A sharp upper bound on discrete Young's inequality for sum with $f$ and $f^{-1}$ 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.
 A: 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 $99/10 - 18\epsilon$.
