Without using the Second Derivative Test, can the convexity of the natural exponential function be shown directly from the definition of convexity? The expression \begin{equation*} e^{t} = \sum_{n=0}^{\infty} \frac{t^{n}}{n!} \end{equation*} can be used. Here is what I have.
The natural exponential function is convex if, and only if, for any pair of conjugate real numbers $s$ and $t$ and for any real numbers $x$ and $y$, \begin{equation*} se^{x} + te^{y} \geq e^{sx + ty} . \end{equation*} This latter inequality is equivalent to \begin{equation*} \sum_{n=0}^{\infty} \frac{sx^{n} + ty^{n}}{n!} \geq \sum_{n=0}^{\infty} \frac{(sx + ty)^{n}}{n!} \end{equation*} and to \begin{equation*} \sum_{n=0}^{\infty} \frac{sx^{n} + ty^{n} - (sx + ty)^{n}}{n!} \geq 0 . \end{equation*} Since $s + t = 1$, this last inequality is equivalent to \begin{equation*} \sum_{n=2}^{\infty} \left( s(1 - s^{n-1})x^{n} + t(1 - t^{n-1})y^{n} - \frac{1}{n!} \sum_{i=1}^{n-1} \binom{n}{i} (sx)^{i}(ty)^{n-i}\right) \geq 0 . \end{equation*}
I am not sure that this last inequality is useful.
