Exponential type inequality I found this problem in a set of old analysis prelims.  I am looking for some hints after having spent some time working on the problem.
Suppose $a_1,\ldots,a_n$ are real numbers such that $1+a_j>0$ for all $1\leq j\leq n$.
If $\sigma_n=\frac{a_1}{1+a_1}+\ldots +\frac{a_n}{1+a_n}\geq0$, then
$$
1+\frac{\sigma_n}{1!}+\ldots + \frac{\sigma^n_n}{n!}\leq \prod^n_{j=1}(1+a_j)
$$
with equality only if $a_j=0$ form all $j$.
I tried to apply common inequalities such as $1+x\leq e^x$ but the problem is still elusive. A hint or a solution would be welcome.
 A: Hint 1: Strengthen the inequality to make it variable-seperable:
$$e ^ {\sigma_n} = \sum_{i=0}^\infty \frac{ \sigma_n ^ i } { i! } \leq \prod ( 1 + a_i).$$
This uses $ \sigma_n \geq 0 $.
Hint 2: Show that $ e^ \frac{a_i } { 1 + a_i }  \leq 1 + a_i$ on $ a_i > -1$.

 From your common inequality with $ x = \frac{-a_i } {  1 + a_i }$, we get $ e ^ {\frac{-a_i } {  1 + a_i }}  \geq 1 + \frac{-a_i } {  1 + a_i } = \frac{1}{ 1 + a_i } $.
 Cross multiply (verify terms are positive) to get the desired inequality.
 Equality holds when $ x = 0$, so $ a_i = 0$.

A: Notice that for any $\sigma\geq0$
$$\begin{align}
1+\frac{\sigma}{1!}+\ldots +\frac{\sigma^n}{n!}\leq e^\sigma\tag{1}\label{one}
\end{align}$$
with equality only if $\sigma=0$. From this you get
$$
1+\frac{\sigma_n}{1!}+\ldots +\frac{\sigma^n_n}{n!}\leq \prod_{1\leq l\leq n}\exp\big(\tfrac{a_j}{1+a_j}\big)
$$
The following inequality will complete the job: $e^{\tfrac{x}{1+x}}\leq 1+x$, which is valid for $1+x>0$, and with equality only if $x=0$. This inequality may be derived from the other inequality:
$$\begin{align}
u\log(u)&\geq u-1,\qquad u\geq0\tag{2}\label{two}\end{align}$$
by setting $u=1+x$. To get \eqref{two}, you can use simple differential calculus.

 Let $\phi(u)=u\log(u)-u$. Then $\phi'(u)=\log(u)$ has a unique root at $u=1$. Since $\phi''(u)=\frac1u>0$, it follows that $u=1$ is a local minimum point. Since $\phi(0)=0$ and $\lim_{u\rightarrow\infty}\phi(u)=\infty$, $u=1$ is a global minimum of $\phi$ in $[0,\infty)$, and \eqref{two} follows.

