Let $x_1,x_2,...x_n$ be $n$ positive numbers such that their product is equal to $k$. Find min value of $f(x)=(1+x_1)(1+x_2)...(1+x_n)$ I am trying to solve a physics problem which reduces to the above problem.I tried approaching the above problem using AM-GM inequality and arrived at the following results.
$$\frac{(x_1 + x_2+x_3 +  ....x_n)}{n}\geq k^\frac{1}{n}$$
$$\frac{(x_1 + x_2+x_3 +  ....x_n)}{n}\geq f(x)^\frac{1}{n}-1$$
 A: Consider the function $f : \mathbb{R} \to \mathbb{R}$, $t \mapsto \ln(1+e^t)$.  Then $f$ is a convex function since $f''(t) = \frac{e^t}{(1+e^t)^2}$ for all $t$.  It follows that for all $t_1, \ldots, t_n \in \mathbb{R}$:
$$f \left( \frac{t_1 + \cdots + t_n}{n} \right) \le \frac{f(t_1) + \cdots + f(t_n)}{n}.$$
Plugging in $t_i := \ln x_i$, we get:
$$\ln (1 + e^{(\ln x_1 + \cdots + \ln x_n)/n}) \le \frac{\ln(1+x_1) + \cdots + \ln(1+x_n)}{n}.$$
Since $x_1 \cdots x_n = k$, we have $\ln x_1 + \cdots + \ln x_n = \ln k$, so
$$\ln(1 + k^{1/n}) = \ln(1 + e^{(\ln k)/n}) \le \frac{1}{n} \ln [(1+x_1) \cdots (1+x_n)].$$
Rearranging to solve for $(1+x_1)\cdots(1+x_n)$ finally gives:
$$(1+x_1) \cdots (1+x_n) \ge (1+k^{1/n})^n.$$
(Since $f$ is actually strictly convex, it is also easy to see that we have equality if and only if $x_1 = \cdots = x_n = k^{1/n}$.)
A: A simple upper bound for $\min f$ is obtained by letting all $x_i$ eequal to $\sqrt[n]k$, i.e., certainly $\min f\le (1+\sqrt[n]k)^n$.
As all factors in $f$ are $\ge 1$, we can ignore all cases where any $x_i$ is $>b:=(1+\sqrt[n]k)^n-1$. Thus we may also assume that each $x_i$ is $\ge a:=\frac{k}{b^{n-1}}>0$.
On the compact set $\{\,x\mid a\le x_i\le b, x_1\cdots x_n=k\,\}$, the continuous function $f$ attains its minimum (so we are not chasing a phantom).
Let the minimum be attained at some point $(x_1,\ldots,x_n)$.
Assume that $x_i\ne x_j$ for some $i,j$.
Then 
$$(1+x_i)(1+x_j)=1+x_ix_j+x_i+x_j\ge 1+x_ix_j+2\sqrt{x_ix_j} $$
with equality iff $x-i=x_j$. Thus if $x_i\ne x_j$, we can decrease the value of $f$ by replacing $x_i$ and $x_j$ with their geometric mean.
By assumption, no such decrease is possible. We conclude that $x_i=x_j$ - for all $i,j$.
A: You can reduce problem to the case $n=2$.
 Assume at minimum point you have $x_i =x^*_i$ for all $i \in \{3,4,...,n\}$
Then you are going to minimize $f(x_1 , x_2)= \alpha (1+x_1)(1+x_2)$ subject to $x_1 x_2 = \frac{k}{\lambda}. $ Where $\lambda , \alpha > 0 $. By solving the later problem you get $x_1 = x_2 = \sqrt \frac{k}{\lambda}. $ Now this tells you (inductively) $x_1 = x_2 = ... = x_n$ where minimum is attained.
