cursed_tangent1434 wrote:

Posting for storage. Consider a single bracketed term.
$(1+a_n)^n$ In such a term, after expanding and counting multiple like terms separately we will have $2^n$ terms. Then applying AM-GM we see that,
$\begin{align*} (1+a_n)^n &= a_n^n+na_n^{n-1}+\dots + 1\\ &\geq 2^n(\sqrt[2^n]{a_n^{k}})\\ &\geq 2^n(a_n^{\frac{k}{2^n}})\\ & > 2^na_n \end{align*}$ Here, notice that we obtain the fourth line from $k> 2^n$ ( $k=2^n$ when $n=2$ ) since we are multiplying the coefficient of each term with the power to obtain $k$ , but $2^n$ is simply the sum of the coefficients.
Then, notice that the $LHS$ of the required inequality, $S$
$\begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}} \end{align*}$ This reduces the desired inequality to $2^{\frac{n^2+n-2}{2}} \geq n^n$
Claim : $2^{\frac{n^2+n-2}{2}} \geq n^n$ for all $n \geq 2$

Proof : Now, notice that
$\begin{align*} &2^{\frac{n^2+n-2}{2}}\geq n^n\\ &\Longleftrightarrow 2^{\frac{n^2+n-2}{2n}}\geq n \end{align*}$ This inequality is obviously true for $n=2$ (This is when we have equality). Then, if we have $2^{\frac{n^2+3n}{2n+2}-\frac{n^2+n-2}{2n}}\geq 1$ for all $n$ , then the required inequality (second line) will always be true. It is easy to verify this (the LHS is much larger than 1 when $n$ gets larger).

Now, notice that this means,
$\begin{align*} S &> 2^2\cdot 2^3 \dots 2^n a_2a_3\cdots a_n\\ &= 2^{\frac{n^2+n-2}{2}}\\ &\geq n^n \end{align*}$ which was the required inequality.

splendid solution! :first: