Prove that $E(e^{X}) < \frac{1}{2}(e^{-K} + e^{K})$ if $E(X)=0$ and $|X|<K$ almost surely

If $X$ only takes values in $[-K,K]$ then it is true, for all convex $\phi$, that $\phi(X) \leqslant \frac{(\phi(K)-\phi(-K))}{2K}X+\frac{\phi(-K)+\phi(K)}{2}$. This is because the right hand is the equation of the line from $(-K,\phi(-K))$ to $(K,\phi(K))$

Setting $\phi=\exp$ and taking expectations in the above inequality gives $\mathbb{E}(e^X)\leqslant \frac{1}{2}(e^{K}+e^{-K})$

To get the strict inequality, notice that if $\phi$ is strictly convex and $X$ takes values in $(-K,K)$ all inequalities above become strict. (Strict inequalities stay strict by taking expectations because positive variables have positive expectations)

Jensen's inequality's proof looks a lot like the argument above, but gives a minorant of $\phi(X)$ by a line, instead of a majorant like it is the case here.