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

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.