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EDIT: This is wrong -- careless mistake as noted in the comments. I thought I had deleted it, but here it still is.

Working with the RHS of your inequality we have

\begin{eqnarray}\sum_i (P_i - Q_i) \log{\left(\frac{1}{\frac{P_i}{e} + (1-\frac{1}{e})Q_i}\right)} &=& \sum_i (P_i - Q_i)\log{\left(\frac{e}{P_i + (e-1)Q_i}\right)}\\ & = & \sum_i (P_i - Q_i) (1 - \log{(P_i + (e-1)Q_i)})\\ & = & \sum_i (P_i - Q_i) + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}\\ & = & 1 - 1 + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}\\\ & = & \sum_i Q_i \log{(P_i + (e-1)Q_i)} - \sum_i P_i \log{(P_i + (e-1)Q_i)}\\\ & \geq & \sum Q_i \log{(Q_i)} - \sum_i P_i \log{(P_i)}\\ & =& -\mbox{H}(Q) + \mbox{H}(P). \end{eqnarray} The inequality follows from $\log{(P_i)} \leq \log{(P_i + (e-1)Q_i)}$ and $\log{(Q_i)} \leq \log{(P_i + (e-1)Q_i)}$.

Post Deleted by R Hahn
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Rewriting $$\log{\left(\frac{1}{\frac{P_i}{e} Working with the RHS of your inequality we have \begin{eqnarray}\sum_i (P_i - Q_i) \log{\left(\frac{1}{\frac{P_i}{e} + (1-\frac{1}{e})Q_i}\right)} = &=& \log{\left(\frac{e}{P_i sum_i (P_i - Q_i)\log{\left(\frac{e}{P_i + (e-1)Q_i}\right)} e-1)Q_i}\right)}\\ & = & \sum_i (P_i - Q_i) (1 - \log{(P_i + (e-1)Q_i)},$$ your rhs simplifies as $$\sum_i e-1)Q_i)})\\ & = & \sum_i (P_i - Q_i) + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}.$$

Because e-1)Q_i)}\\ & = & 1 - 1 + \sum_i (Q_i - P_i)\log{(P_i + (e-1)Q_i)}\\\ & = & \sum_i Q_i \log{(P_i + (e-1)Q_i)} - \sum_i P_i \log{(P_i + (e-1)Q_i)}\\\ & \geq & \sum Q_i \log{(Q_i)} - \sum_i P_i \log{(P_i)}\\ & =& -\mbox{H}(Q) + \mbox{H}(P). \end{eqnarray} The inequality follows from $\log{(P_i)} \leq \log{(P_i + (e-1)Q_i)}$ and $\log{(Q_i)} \leq \log{(P_i + (e-1)Q_i)}$ the inequality follows. e-1)Q_i)}\$.

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