Sum of vector element product - product of sum inequality I'm looking for the name (and proof) of the following inequality:
Given vector $\bf{a}$ and $\bf{b}$:
$$
\Pi_i a_i + \Pi_ib_i \leq \Pi_i(a_i+b_i)
$$
This is probably trivial and well-known, but I can't remember where I've seen this. Thanks!
 A: Let's see how far we get without extra hypotheses. 
For $n=1$ there is nothing to do, for $n=2$ we have
$$
0 \le a_1 b_2 + a_2 b_1
$$
which already makes clear that there is no unique answer. Let's exploit this for general $n$.
Equivalent statements of the problem are for $n \ge  2$:
$$
\Pi_i a_i + \Pi_ib_i \leq \Pi_i(a_i+b_i) \leftrightarrow \\
\Pi_i a_i (1 + \Pi_ix_i) \leq \Pi_i a_i \Pi_i(1+x_i)
$$
where $x_i = b_i/a_i$. 
Now 
$$
\Pi_i(1+x_i) = 1 + \sum_i x_i + \sum_{1\leq i<j\leq n} x_i x_j + \cdots + \Pi_ix_i \sum_j \frac{1}{x_j} + \Pi_ix_i
$$
which gives us 
$$
0 \le \Pi_i a_i (\sum_i x_i + \sum_{1\leq i<j\leq n} x_i x_j + \cdots + \Pi_ix_i \sum_j \frac{1}{x_j})
$$
In these formulations, it is understood that there will be no undefined terms once $\Pi_i a_i $ is multiplied into the bracket again.
If all $a_i \ge 0$ and $b_i \ge 0$ this proves the inequality to hold.
If all  $a_i \le 0$ and $b_i \le 0$ then we can also draw some conclusions. As in this case, all $x_i \ge 0$, the last bracket is $ \ge 0$, so the inequality holds if also  $\Pi_i a_i \ge 0 $ which will be the case for even $n$. 
Making this explicit, the lowest meaningful $n$ for which this happens is $n=2$ where we have already seen 
$$
0 \le a_1 b_2 + a_2 b_1
$$
which is true since all variables are nonpositive.
For $n=3$, we would have
$$
0 \le a_1 a_2 b_3 + a_1 b_2 b_3 + a_1 b_2 a_3 + b_1 a_2 a_3 + b_1 b_2 a_3 + b_1 a_2 b_3 
$$
which can be falsified by, for example, setting all variables to $-1$.
Other choices of the signs of  the variables give no clear results.
