Inequality deduced from martinagles 
Let $c$ be a positive real constant, and let $x_i,i \in \{1,2,...,n\}$  be real numbers such that $$  |x_i|\le c,\forall i \in \{1,2,...,n\}.$$ 
Let $p_i,i \in \{1,2,...,n\}$ be positive reals such that  $$\sum_{i=1}^n p_i x_i=0.\\$$ $$\sum_{i=1}^n p_i=1$$
  Prove that 
$$\sum_{i=1}^n {p_i} e^{\lambda x_i}\le e^{\frac {{\lambda}^2 c^2} {2}}.$$

I have a proof using Taylor expansion, is there a simple proof?
 A: Ok this problem can be seen from the perspective of mean preserving spreads. Take some convex function $f$. additionally take a Random variable $X$ and a random variable $Y=X+\epsilon$ where $\epsilon$ is mean zero. When $Y$ can be written this way we say that $Y$ is a mean preserving spread of $X$ Then we know that, $$E[f(X)]<E[f(Y)]$$ This is proved in Blackwell (1953). So to apply this to the current question we want to prove the inequality for all RVs X whose support is in $[-c,c]$ and who have mean $0$. Well then every RV in the set in a mean preserving transformation of the others. But also note that the left hand side of the inequality is equal to $E[f(x)]$ where $f=e^{\lambda x}$ is a convex function.
This means that that performing a mean preserving spread will increase the LHS. Thus we only have to prove the inequality for the distribution that is most spread and this will suffice for all other distributions. But the most spread distribution is 1/2 mass on $c$ and 1/2 mass on $-c$, and so the inequality reduces to $$\frac{e^{\lambda c}+e^{-\lambda c}}{2}\leq e^{\frac {{\lambda}^2 c^2} {2}}$$
At this point you can come as close to a proof by graphing as one can. Because the inequality is obvious when $\lambda c\geq 1$ and so you can graph it between $[0,1]$ and see it hold.
Alternatively you can prove it by continuous induction by using a second order taylor approximation. This might have been what you were talking about, but the first step probably makes it a lot easier if you didn't do that already.
Also, one more note. If $\lambda$ is negative then it holds trivially because then the function is concave and spreads decrease the expectation. But this means that the LHS is maximized at the degenerate distribution $x_i=0 \;\forall i$. And so the LHS is maxed at 1.
Also the above value, $$\frac{e^{\lambda c}+e^{-\lambda c}}{2}$$ is a better bound.
