As an exercise, I thought I would try to prove some classical Chernoff bounds without ever using the moment generating function, but then found myself getting stuck in certain places. Before I state my question, let me give some background.

$\mathbf{Background}$:

Consider for example the fact that there exist constants $c,c'$ such that for any $\lambda > 0$ \begin{equation} (*)\ \mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) \le c \cdot e^{-c' \lambda^2/\|x\|_2^2} \end{equation} where $\sigma_1,\sigma_2,\ldots$ are independent Rademachers. One way to prove (*) for small $\lambda$ is to say $$ \mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) = \mathop{\mathbb{P}}_\sigma\left(e^{t\sum_i \sigma_i x_i} > e^{t\lambda}\right) < e^{-t\lambda} \cdot \mathop{\mathbb{E}}_\sigma e^{t\sum_i \sigma_i x_i} = e^{-t\lambda}\cdot \prod_i \mathop{\mathbb{E}}_{\sigma_i} e^{t\sigma_i x_i} = e^{-t\lambda}\cdot \prod_i \frac 12(e^{-tx_i} + e^{tx_i}) \le e^{-t\lambda} \prod_i \frac 12(1 + e^{tx_i}t^2x_i^2/2) \le e^{-t\lambda} \prod_i e^{e^{tx_i}t^2x_i^2/2} \le e^{c\cdot t^2\|x\|_2^2/2 - t\lambda} $$ as long as $t$ (which will depend on $\lambda$) is small enough so that the $e^{tx_i}$ term is at most $c$. Then one optimizes and sets something like $t \sim \lambda/\|x\|_2^2$. This proof uses the moment generating function (MGF), i.e. we talk about $\mathbb{E} e^{tX}$ in the proof for our random variable $X = \sum_i \sigma_i x_i$. At various points in the proof we also use analytic properties of the exponential function, e.g. based on Taylor's theorem.

A proof that avoids the MGF and Taylor's theorem is to work with moments. Define $\|X\|_p = (\mathbb{E}|X|^p)^{1/p}$ as usual and let $g_1,g_2,\ldots$ be i.i.d. gaussians of mean $0$ and variance $1$. Then $$ \|\sum_i \sigma_i x_i\|_p = \sqrt{\frac{\pi}2} \|\mathop{\mathbb{E}}_g \sum_i \sigma_i |g|_i x_i\|_p \le \sqrt{\frac{\pi}2} \|\sum_i \sigma_i |g_i| x_i\|_p = \sqrt{\frac{\pi}2} \|\sum_i g_i x_i\|_p \lesssim \|x\|_2\cdot \sqrt{p} $$ where the last inequality used $2$-stability of the gaussian (also the $\sqrt{\pi/2}$ can be avoided but never mind about that). Then we use that $$ \mathop{\mathbb{P}}_\sigma\left(\sum_i \sigma_i x_i > \lambda\right) \le \mathop{\mathbb{P}}_\sigma\left(|\sum_i \sigma_i x_i|^p > \lambda^p\right) < \lambda^{-p} \cdot \|\sum_i \sigma_i x_i\|_p^p $$ and set $p\sim \lambda^2 / \|x\|_2^2$. This proof is of the form I'm looking for: no MGF, no analytic tricks/Taylor's theorem. (It also works for all $\lambda$.)

$\mathbf{My\ question}$: One formulation of the Chernoff bound is the following. There is some constant $c>0$ such that the following holds. Say $X_1,\ldots,X_n$ are independent and each bounded by $K$ in magnitude almost surely. Let $X = \sum_i X_i$ with $\mu = \mathbb{E} X$ and $\sigma^2 = \mathbb{E}(X - \mathbb{E}X)^2$. Then for all $\lambda > 0$, $$ (**)\ \mathop{\mathbb{P}}\left(|X - \mu| > \lambda\right) \lesssim \max\left\{e^{-c\lambda^2/\sigma^2}, \left(\frac{\sigma^2}{\lambda K}\right)^{c\lambda/K}\right\} . $$ One way to prove the above is to use the moment generating function and analytic tricks (see Exercise 3 at http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/). My question is, similar to the first example, can we prove (**) simply using moment methods while avoiding the MGF and Taylor's theorem/analytic tricks? e.g. using combinations of Jensen's inequality, symmetrization, triangle inequality on $\|\cdot \|_p$, and other such things.

Note that (**) seems to be equivalent to the statement that for all $p \ge 1$, $$ \|X - \mu\|_p \lesssim \sigma\sqrt{p} + K\frac{p}{\ln(epK^2/\sigma^2)} . $$