A Hamiltonian inequality using uniform convexity and Taylor's formula (Evans PDE, §3.3.3, Lemma 4, (36)) In the proof of Lemma 4 (Semiconcavity again) in section 3.3.3 of the book "Partial differential equations" written by Lawrence C. Evans, there's a claim

We note first using Taylor's formula that (35) implies 
(36)       $ H(\frac{p_1+p_2}{2}) \leq \frac{1}{2}H(p_1)+\frac{1}{2}H(p_2)-\frac{\theta}{8}|p_1-p_2|^2 $.

How does uniform convexity and Taylor's formula imply this?
I know that by Taylor's formula, 
$ H(\frac{p_1+p_2}{2}) = H(p_1)+DH(p_1)\cdot(\frac{p_2-p_1}{2})+(\frac{p_2-p_1}{2})^TD^2H(p_1)(\frac{p_2-p_1}{2})  $, 
and 
$ H(\frac{p_1+p_2}{2}) = H(p_2)+DH(p_2)\cdot(\frac{p_1-p_2}{2})+(\frac{p_1-p_2}{2})^TD^2H(p_2)(\frac{p_1-p_2}{2})  $, 
add them and then divide by 2, we get 
$ H(\frac{p_1+p_2}{2}) = \frac{H(p_1)}{2}+\frac{H(p_2)}{2}+(\frac{DH(p_2)-DH(p_1)}{2})\cdot(\frac{p_1-p_2}{2})+(\frac{p_1-p_2}{2})^T(\frac{D^2H(p_2)-D^2H(p_1)}{2})(\frac{p_1-p_2}{2})  $. 
But I can NOT get the result.
 A: A couple of mistakes of execution first: (1) Your invocation of Taylor's theorem isn't correct – the second-order term needs be an error term since we want an exact equality, and should have an additional factor of $\frac{1}{k!} = \frac{1}{2}$; (2) The sign of “$D^2 H(p_1)$” (in quotes as by (1) we shouldn't evaluate at $p_1$) in the last equation should be $+$, because double negatives.
Even with those corrections, we have an unwanted nonzero first-order term, and a sign on the second-order term that's opposite of what we want. But: we can resolve these issues by applying Taylor at $\boldsymbol{\frac{p_1+p_2}{2}}$ instead!
Proof. Let $\xi = \frac{p_1-p_2}{2}$. By Taylor ($C^2$ with Lagrange remainders) there exist points $e_1, e_2$ on the line segment $\overline{p_1p_2}$ such that
$$
\begin{align*}
H(p_1) &= H\left(\frac{p_1+p_2}{2}\right) + \sum_{i=1}^n H_i\left(\frac{p_1+p_2}{2}\right) \xi_i + \frac{1}{2} \sum_{i,j=1}^n H_{ij}(e_1) \xi_i \xi_j,
\\
H(p_2) &= H\left(\frac{p_1+p_2}{2}\right) + \sum_{i=1}^n H_i\left(\frac{p_1+p_2}{2}\right) (-\xi)_i + \frac{1}{2} \sum_{i,j=1}^n H_{ij}(e_2) (-\xi)_i (-\xi)_j.
\end{align*}
$$
Averaging the two equations and rearranging, then applying uniform convexity,
$$
\begin{align*}
H\left(\frac{p_1+p_2}{2}\right)
&= \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - \frac{1}{4} \sum_{i,j} H_{ij}(e_1) \xi_i \xi_j - \frac{1}{4} \sum_{i,j} H_{ij}(e_2) \xi_i \xi_j
\\ 
&\leq \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - 2 \cdot \frac{\theta}{4} \lvert \xi \rvert^2
\\
&= \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - \frac{\theta}{8} \lvert p_1 - p_2 \rvert^2. \tag*{$\blacksquare$}
\end{align*}
$$
A: Only a small correction to the answer of @epimorphic. $ e_1 $ is NOT on the segment between $p_1$ and $p_2$ . Because $e_1 = \frac {p_1 + p_2 } {2} + \delta_1 \cdot \xi = \frac {p_1 + p_2 } {2} + \delta_1 \cdot \frac {p_1 - p_2 } {2} $  where $ 0<=\delta<=1 $ , so $e_1$ is on the line segment between $\frac {p_1 + p_2 } {2} $ and $ \frac {p_1 - p_2 } {2}$ . The similarity is in the case of $e_2$.
