Problem: Call a real-valued function $f$ \emph{very convex} if
\[ \frac{f(x)+f(y)}{2}
  \ge f\left( \frac{x+y}{2} \right) + \left\lvert x-y \right\rvert  \]
holds for all real numbers $x$ and $y$.
Prove that no very convex function exists.

Solution: For $C \ge 0$, we say a function $f$ is \emph{$C$-convex}
\[ \frac{f(x)+f(y)}{2}
  \ge f\left( \frac{x+y}{2} \right) + C\left\lvert x-y \right\rvert. \]

Suppose $f$ is $C$-convex.
Let $a < b < c < d < e$ be any arithmetic progression,
such that $t = |e-a|$.
Observe that
\begin{align*}
  f(a) + f(c) &\ge 2f(b) + C \cdot \half t \\
  f(c) + f(e) &\ge 2f(d) + C \cdot \half t \\
  f(b) + f(d) &\ge 2f(c) + C \cdot \half t
\end{align*}
Adding the first two to twice the third gives
\[ f(a) + f(e) \ge 2f(c) + 2C \cdot t. \]
So we conclude $C$-convex function is also $2C$-convex.
This is clearly not okay for $C > 0$.