Proof of a general version of Chebyshev's inequality The question is as follows,

Let $f$ be a non-negative even function that is non-decreasing for positive $x$. Then for a random variable $ξ$ with $|ξ|≤C$, show $\frac{{\mathbb{E}f\left( \xi  \right) - f\left( \varepsilon  \right)}}{{{f(C)}}} \leqslant \mathbb{P}\left\{ {\left| {\xi  - \mathbb{E}\xi } \right| \geqslant \varepsilon } \right\} \leqslant \frac{{\mathbb{E}f\left( {\xi  - \mathbb{E}\xi } \right)}}{{f\left( \varepsilon  \right)}}$.

Can anyone provide some help? Thank you!
 A: For the right inequality, notice that
\begin{align}
& E[f(\xi - E(\xi))] = \int_{|\xi - E(\xi)| \geq \varepsilon} f(\xi - E(\xi)) dP + \int_{|\xi - E(\xi)| < \varepsilon} f(\xi - E(\xi)) dP \\
\geq & \int_{|\xi - E(\xi)| \geq \varepsilon} f(\xi - E(\xi)) dP \quad \text{since $f$ is nonnegative} \\
\geq & \int_{|\xi - E(\xi)| \geq \varepsilon} f(\varepsilon) dP \quad \text{since $f$ is non-decreasing} \\
= & f(\varepsilon)P(|\xi - E(\xi)| \geq \varepsilon).
\end{align}
The left inequality is not correct. For a counterexample, consider 
$$C = 1/4, \; \xi \sim \text{Uniform}(-1/4, 1/4), \; f(x) = |x|, \; \varepsilon = 1/16.$$
Then 
\begin{align*}
E[f(\xi)] - f(\varepsilon) = E[|\xi|] - \frac{1}{16} = \frac{1}{16},
\end{align*}
whereas
\begin{align*}
C^2 P(|\xi| \geq 1/16) = \frac{1}{16}\times \frac{3}{4} < \frac{1}{16}.
\end{align*}

For the corrected inequality, also begin with by splitting $E[f(\xi)]$:
\begin{align}
& E[f(\xi)] = \int_{|\xi - E(\xi)| \geq \varepsilon} f(\xi) dP + \int_{|\xi - E(\xi)| < \varepsilon} f(\xi) dP \\
\leq & f(C) P(|\xi - E(\xi)| \geq \varepsilon) + \int_{|\xi - E(\xi)| < \varepsilon} f(\xi) dP \quad \text{ since $f$ is even, nondecreasing and $|\xi| \leq C$.} \\
\end{align}
For the latter inequality, notice that $\{|\xi - E(\xi)| < \varepsilon\} \subset \{|\xi| \leq \varepsilon + |E(\xi)|\}$, so that
$$\int_{|\xi - E(\xi)| < \varepsilon} f(\xi) dP \leq f(\varepsilon + |E(\xi)|).$$
So what I can show here is 
$$\frac{E[f(\xi)] - f(\varepsilon + |E(\xi)|)}{f(C)} \leq P(|\xi - E(\xi)| \geq \varepsilon),$$
which is still slightly different from what you gave (they agree when $E(\xi) = 0$.)
