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Let $n \ge 3$ be an integer, and let $a_2, a_3, \ldots , a_n$ be positive real numbers such that $a_2\cdots a_n = 1.$
Prove that \[(1+a_2)^2(1+a_3)^3\cdots (1+a_n)^n > n^n\]

Did you miss $(1+a_1)$ at the front? Just asking.

It seems that there is no $a_1$ thanks for pointing out.

\[(a_k+1)=(a_k+\frac {1}{k-1}+\frac {1}{k-1}+.....+\frac {1}{k-1}) \geq k\sqrt [k] {\frac {a_k}{(k-1)^{k-1}}}\]
[By AM-GM inequality]
hence $(a_k+1)^k \geq \frac {a_k.k^k}{(k-1)^{k-1}} \forall k \geq 2$
so $(a_2+1)^2(a_3+1)^3.........(a^n+1)^n \geq a_2.a_3......a_n.\frac{2^2.3^3.........n^n}{1^1.2^2.........(n-1)^{n-1}}=n^n$
equality iff $a_k=\frac {1}{k-1} \forall k \geq 2$
but then $a_2.a_3........a_n=\frac {1}{1.2.3.......(n-1)}$ which is equal to $1$ iff $n=2$ but it's given that $n \geq 3$ so a contradiction for the equality.

My solution is the same as Tahmid vaia's.So no need to post it again.Really charming problem.

Sorry,but forgot to share one thing. I think the process of expanding the term $(1+a_{k})$ is perhaps known as Cauchy Reverse Technique. I'm not sure,but I think I have read about this type of technique.

Almost similar solution. But Tahmid's is neater.

We have by AM-GM,
\[\begin{align*} (1+a_2)^2&\ge 2^2a_2\\
(1+a_3)^3=(1+\frac{a_3}{2}+\frac{a_3}{2})^3&\ge \frac{3^3}{2^2}a_3^2\\
&\vdots\\
(1+a_n)^n=(1+\frac{a_n}{n-1}+\dots+\frac{a_n}{n-1})^n&\ge \frac{n^n}{(n-1)^{n-1}}a_n^{n-1}\end{align*}\]
Multiplying all these we get $(1+a_2)^2(1+a_3)^3\cdots (1+a_n)^n\ge n^na_2a_3^2\cdots a_n^{n-1}=n^na_2^2a_3^3\cdots a_n^n$, using $a_2\cdots a_n=1$. This is equivalent to
\[(1+\frac{1}{a_2})^2(1+\frac{1}{a_3})^3\cdots (1+\frac{1}{a_n})^n\ge n^n.\]
Now replace $1/a_i$ by $a_i$ and we are done. Equality holds iff $a_i=i-1$, impossible. So the inequality is strict.

Yess .. Same As Tahmiad Vai

By inequality Holder we have
$$(1+1)^1(1+a_2)^2(1+a_3)^3...(1+a_n)^n \ge (1+a_2.a_3...a_n)^{\frac{n(n+1)}{2}}$$
$$=2^{\frac{n(n+1)}{2}} = \left( 2^{\frac{(n+1)}{2}}\right)^n \ge ( 2\ln2.n )^n=n^n(2\ln2)^n > 2n^n, (n \ge 3)$$
Note $$f(x)=2^{\frac{x+1}{2}}-2\ln2.x \ge 0, \forall x \ge 3$$
So $$(1+a_2)^2(1+a_3)^3...(1+a_n)^n >n^n$$

Thanks. That was my solution.