A Hamiltonian inequality using uniform convexity and Taylor's formula (Evans PDE, §3.3.3, Lemma 4, (36))

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, $$ \begin{align*} H\left(\frac{p_1+p_2}{2}\right) &= \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - \frac{1}{4} \sum_{i,j} H_{ij}(e_1) \xi_i \xi_j - \frac{1}{4} \sum_{i,j} H_{ij}(e_2) \xi_i \xi_j \\ &\leq \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - 2 \cdot \frac{\theta}{4} \lvert \xi \rvert^2 \\ &= \frac{1}{2} H(p_1) + \frac{1}{2} H(p_2) - \frac{\theta}{8} \lvert p_1 - p_2 \rvert^2. \tag*{$\blacksquare$} \end{align*} $$