Convexity of the natural exponential fuction - directly from the definition 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.
 A: Demonstration in the case that $x > 0$, $y > 0$, and $s, \, t \in \mathbb{Q}^{+}$
Since $s$ is a positive, proper, rational number, there are positive integers $j < k$ such that $s = j/k$, and since $s$ and $t$ are a pair of conjugate numbers, $t = (k - j)/k$.  For any positive, real numbers $z_{1} , \, z_{2} , \, \ldots \, z_{n}$, and for any positive integer $n$, 
\begin{equation*}
\left(\frac{z_{1} + z_{2} + \ldots + z_{k}}{k}\right)^{n}
\leq \frac{{z_{1}}^{n} + {z_{2}}^{n} + \ldots + {z_{k}}^{n}}{k} .
\end{equation*}
If $z_{1} = z_{2} = \ldots = z_{j} = x$ and $z_{j+1} = z_{j+2} = \ldots = z_{k} = y$,
\begin{align*}
(sx + ty)^{n}
&= \left(\left(\frac{j}{k} \right) x + \left(\frac{k - j}{k} \right) y\right)^{n} \\
&= \left(\frac{jx + (k-j)y}{k}\right)^{n} \\
&\leq \frac{jx^{n} + (k-j)y^{n}}{k} \\
&= \left(\frac{j}{k}\right)x^{n} + \left(\frac{k-j}{k}\right)y^{n} \\
&= sx^{n} + ty^{n} .
\end{align*}
Consequently,
\begin{equation*}
\frac{(sx + ty)^{n}}{n!} \leq \frac{sx^{n} + ty^{n}}{n!} .
\end{equation*}
