How to prove/verify the following inequality consisting of polynomial and exponential holds? How to verify the inequation $C^{H}e^{-\lambda^2 C} \leq \delta$ holds for $C = \frac{1}{\lambda^{2}} (2H log \frac{H}{\lambda^2} + log\frac{1}{\delta})$? Namely, we treat $H \geq 1$, $\lambda > 0$, and $\delta > 0$ as constants and $C$ as the variable. 
Furthermore, can we provide a general condition of C under which the inequation will hold? 
To provide some context, this inequation is a simplified version of Eq (17) in paper A Sparse Sampling Algorithm for Near-Optimal
Planning in Large Markov Decision Processes with $V_{max} = 1$ and $K=1$. By proving this inequation, we are able to show Theorem 1 in that paper is correct and thus the "sparse sampling algorithm" is indead "near-optimal". 
Thanks. 
 A: I assume that $\log$ means $\log_e$. Then my calculations shows that the claim can be wrong. Indeed,
$$C^{H}e^{-\lambda^2 C}=\left(\frac{1}{\lambda^{2}}\left(2H \log \frac{H}{\lambda^2} + \log\frac{1}{\delta}\right)\right)^H e^{-2H\log \frac{H}{\lambda^2} - \log\frac{1}{\delta}}=$$
$$\left(\frac{1}{\lambda^{2}}\left(2H \log \frac{H}{\lambda^2} + \log\frac{1}{\delta}\right)\right)^H \left(\frac{H}{\lambda^2}\right)^{-2H}\delta=$$
$$\left(\frac{\lambda^2}{H^2}\left(2H \log \frac{H}{\lambda^2} + \log\frac{1}{\delta}\right)\right)^H\delta.$$
If $H=\lambda^2$ then $C^{H}e^{-\lambda^2 C}=(\log\frac 1{\delta})^H\delta$. It can be bigger than $\delta$, for instance, when $H=1$ and $\delta=\frac 1{e^2}$.  

can we provide a general condition of C under which the inequation will hold?

Since $0^He^{-\lambda^2 0}=0$, by the continuity of the left hand side, the inequality $C^{H}e^{-\lambda^2 C}\le\delta$ holds for sufficiently small $C$. To satisfy the inequality it suffices to take any $0<C\le 1$ such that $ e^{-\lambda^2 C}\le\delta$, that is $- C\le\frac{\log\delta}{\lambda^2}$.
