Is $\left(\sum_{n=1}^N\frac{a_n}{N}\right)^N\left(\sum_{n=N+1}^{2N}\frac{a_n}{N}\right)^N≠\left(\sum_{n=1}^{2N}\frac{a_n}{2N}\right)^{2N}$?

Question

Let $$G_N= \prod_{n=1}^Na_n$$ and $$A_N=\left(\frac{\sum_{n=1}^Na_n}{N}\right)$$ So $$G_{2N}= \prod_{n=1}^{2N}a_n \\ =\left(\prod_{n=1}^{N}a_n\right)\left(\prod_{n=N+1}^{2N}a_n\right) \\ ≤_{IH}\left(\sum_{n=1}^N\frac{a_n}{N}\right)^N\left(\sum_{n=N+1}^{2N}\frac{a_n}{N}\right)^N$$

I failed to understand why the following is true

$$\left(\sum_{n=1}^N\frac{a_n}{N}\right)^N\left(\sum_{n=N+1}^{2N}\frac{a_n}{N}\right)^N$$

does it imply

$$\left(\sum_{n=1}^{2N}\frac{a_n}{2N}\right)^{2N}$$

But I use N=10 as example with Mathematica

$$\left(\sum_{n=1}^N\frac{a_n}{N}\right)^N\left(\sum_{n=N+1}^{2N}\frac{a_n}{N}\right)^N≠\left(\sum_{n=1}^{2N}\frac{a_n}{2N}\right)^{2N}$$

See the following image

Answer 1

This, does not imply an equality, but an inequality, with $(A_{2N})^{2N}$, as by the two variable AM-GM, we have $$\bigg(\sum_{n=1}^N \frac{a_n}{N}\bigg)\bigg(\sum_{n=N+1}^{2N}\frac{a_n}{N}\bigg) \leq \bigg(\sum_{n=1}^{2N}\frac{a_n}{2N}\bigg)^2 $$ Thus we have $$G_{2N} \leq (A_{2N})^{2N}$$ You can figure out how the equality case transfers. Using induction, this suffices as a proof for all AM-GM inequalities with $2^k$ variables, which is what I believe whatever you're reading is attempting to get at.

Answer 2

To show that your equality is false, all that is needed is an example.

Take $a_1 = 1$ and $a_k = 0$ for $k > 1$.

Then $\left(\sum_{n=1}^N\frac{a_n}{N}\right)^N\left(\sum_{n=N+1}^{2N}\frac{a_n}{N}\right)^N =0 $ and $\left(\sum_{n=1}^{2N}\frac{a_n}{2N}\right)^{2N} =\frac1{(2N)^{2N}} $.