Minimum of $\prod_{1\le i\le n} (1+a_i)$ when $a_1a_2\cdots a_k = M.$ I asked a question here, and also got its generalization:(see tc1729's answer)

$$\prod_{1\le i\le n} (1+a_i)\ge 2^{n}\sqrt{a_1a_2\cdots a_n}=2^n\cdot\sqrt M,$$for$$a_1a_2\cdots a_k = M.$$

But I can't understand that how $$\prod_{1\le i\le n} (1+a_i)$$ attends minimum when $$a_1=\cdots = a_n = \sqrt[n]{M}?$$ Also, as I asked here, $$(1+\sqrt[n]{M})^n\neq2^n\cdot\sqrt M.$$So, if $$\prod_{1\le i\le n} (1+a_i)$$ attends its minimum at $$a_1=\cdots = a_n = \sqrt[n]{M},$$ then min value should be $$(1+\sqrt[n]{M})^n$$ and not $$2^n\cdot\sqrt M.$$
Please help. I am confused.
 A: To complement Ewan Delanoy's answer, let me include an AM-GM proof of Huygen's as well. We shall prove the homogenized version (i.e. $x_i = b_i/a_i$):
$$(a_1^n + b_1^n)(a_2^n + b_2^n) \cdots (a_n^n + b_n^n) \ge (a_1a_2 \cdots a_n + b_1b_2 \cdots b_n)^n$$
for $a_1,\cdots,a_n,b_1,\cdots,b_n > 0$.
If we replace $(a_i,b_i)$ by $(ca_i,cb_i)$ for some positive $c$, we see that it's the same inequality, since the inequality is homogeneous in $a_i,b_i$ on both sides. Scale each $(a_i,b_i)$ so that $a_i^n + b_i^n = 1$, then we need to prove
$$1 \ge (a_1a_2\cdots a_n + b_1b_2 \cdots b_n)^n$$
or just
$$1 \ge a_1a_2\cdots a_n + b_1b_2 \cdots b_n$$
under the condition $a_i^n + b_i^n = 1$ for each $i$. This clearly follows from AM-GM.
A: Putting $x_k=a_k^{\frac{1}{n}}$, one can reformulate the inequality to
be shown as
$$
P_n : (1+x_1^n)(1+x_2^n)\ldots (1+x_n^n) \geq (1+x_1x_2\ldots x_n)^n
\ (\text{for any nonnegative} \ x_1,x_2,\ldots ,x_n)
$$
Let us show $P_n$ by induction on $n$. We have that $P_1$ is an equality, 
$P_2$ is easily equivalent to $(x_1-x_2)^2 \geq 0$. Suppose now that $n\geq 3$ and that $P_{n-1}$ holds. Let $x_1,x_2, \ldots x_n$ be nonnegative
numbers. We want  to show that 
$$
(1+x_1^n)(1+x_2^n)\ldots (1+x_n^n) \geq (1+x_1x_2\ldots x_n)^n \tag{1}
$$
If we put $b=(x_2x_3\ldots x_n)^{\frac{n}{n-1}}$, then applying $P_{n-1}$ to the
$n-1$ numbers $x_2,x_3, \ldots x_n$, we see that it suffices to show (1) when
all the $x_i$ are equal to $b$. In other words, we may assume without loss
of generality that $x_2=x_3=\ldots =x_n$, in which case (1) becomes
$$
(1+x_1^n)(1+x_2^n)^{n-1} \geq (1+x_1x_2^{n-1})^n \tag{2}
$$
If we put $y=x_2^n$ and $z=\frac{x_1}{x_2}$, (2) can be rewritten as
$$
(1+yz^n)(1+y)^{n-1} \geq (1+yz)^n \tag{3}
$$
So if we put $\delta=(1+yz^n)(1+y)^{n-1} - (1+yz)^n$, we must show that
$\delta$ is nonnegative. But expanding the Newton binomials leads to
$$
\delta=\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}-
\frac{1-z^n}{n}\bigg) \tag{4}
$$
So all we need to show now is that the sequence $(\frac{1-z^n}{n})$ is decreasing in
$n \geq 1$, for any $z>0$. The algebraic equality 
$$
\frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}=
\frac{(z-1)^2}{n(n+1)}\bigg(
\sum_{k=0}^{n-1} (k+1)z^k
\bigg) \tag{5}
$$
finishes the proof.
