TITLE: Expectation Problem (Khintchine's Inequality)
QUESTION [7 upvotes]: As a reference, this is Problem 6.2 in Albiac and Kalton's Topics in Banach Space Theory
The question involves a direct proof of Khintchine's Inequality.  In part (1), we are to prove that $\cosh(t)\leq e^{t^2/2}$ for all $t\in\mathbb{R}$.  Part (2) (which I don't think I need yet for the step I am stuck on) is to show that for $p\geq1$, $t^p\leq p^pe^{-p}e^t$ for I assume $t>0$.  These steps I have done.
The part I am stuck on is this: Let $(\varepsilon_n)_{n=1}^\infty$ be a sequence of Rademachers (i.e. independent random variables taking on values $\pm1$ with equal probability), and let $(a_k)_{k=1}^\infty\subset\mathbb{R}$ such that $$\underset{k=1}{\overset{n}{\sum}}a_k^2=1$$ and define $$f=\underset{k=1}{\overset{n}{\sum}}a_k\varepsilon_k$$
We want to show that $$\mathbb{E}(e^f)\leq e$$
(If you have any doubts, this is expectation over the epsilons).
Here has been my attempt so far:
$$\mathbb{E}(e^f)=\mathbb{E}\left(\exp\left(\underset{k=1}{\overset{n}{\sum}}a_k\varepsilon_k\right)\right)=\mathbb{E}\left(\underset{k=1}{\overset{n}{\prod}}\exp(a_k\varepsilon_k)\right)=\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}\exp(a_k\varepsilon_k)\right)\leq\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}(2\cosh(a_k\varepsilon_k))\right)$$
the second to last equality is by independence, and the last inequality here follows from the definition of $\cosh$.  Then by part (1), we have
$$\leq\underset{k=1}{\overset{n}{\prod}}\left(\mathbb{E}\left(2\exp\left(\frac{a_k^2\varepsilon_k^2}{2}\right)\right)\right)$$
But since $\varepsilon_k^2=1$ for all $k$, and we can move the expectation back outside, and put the product back into the exponent as a sum, we will get something like $$2^n\exp\left(\underset{k=1}{\overset{n}{\sum}}\frac{a_k^2}{2}\right)=2^n\exp\left(\frac{1}{2}\right)$$ by assumption on the sum of $a_n^2$.  So this does not give the inequality we want.
Any suggestions or hints would be appreciated (be gentle, I am far from knowledgeable of probability theory)

REPLY [6 votes]: I think you should just evaluate $\mathbb{E}\exp(a_k\varepsilon_k)$ instead of bounding it. By definition, we have 
$$\mathbb{E}\exp(a_k\varepsilon_k)=\exp(-a_k)\,\mathbb{P}(\varepsilon_k=-1)+
\exp(a_k)\,\mathbb{P}(\varepsilon_k=1)={\exp(-a_k)+\exp(a_k)\over 2}.$$
This gives $\cosh(a_k)$, and your expectation becomes $\prod\limits_{k=1}^n\cosh(a_k)$.
Now use part (1) of the exercise to show that $\mathbb{E}(e^f)\leq e^{1/2}$.