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To prove the given inequality:

\[ \frac{2n(2n-1)\cdots(n+1)}{n(n-1)\cdots 2 \cdot 1} \geq \frac{4^n}{n+1} \]

for all \( n \in \mathbb{N} \), let's proceed step by step.

First, let's rewrite the left-hand side of the inequality:

\[ \frac{2n(2n-1)\cdots(n+1)}{n(n-1)\cdots 2 \cdot 1} \]

Notice that the numerator is the product of \( n \) consecutive integers starting from \( 2n \) and going down to \( n+1 \), and the denominator is \( n! \). This can be written as:

\[ \frac{(2n)!}{n! \cdot n!} \]

This is the binomial coefficient \( \binom{2n}{n} \), which counts the number of ways to choose \( n \) elements from \( 2n \) elements:

\[ \binom{2n}{n} = \frac{(2n)!}{n! \cdot n!} \]

So, the inequality becomes:

\[ \binom{2n}{n} \geq \frac{4^n}{n+1} \]

Now, we use the fact that \( \binom{2n}{n} \) can be approximated using Stirling's approximation for factorials:

\[ n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \]

Using this approximation, we get:

\[ \binom{2n}{n} \approx \frac{\sqrt{4\pi n} \left(\frac{2n}{e}\right)^{2n}}{\left( \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \right)^2} = \frac{\sqrt{4\pi n} \cdot \left(\frac{2n}{e}\right)^{2n}}{2\pi n \cdot \left(\frac{n}{e}\right)^{2n}} = \frac{\sqrt{4\pi n} \cdot 2^{2n} \cdot n^{2n}}{2\pi n \cdot n^{2n}} = \frac{4^n \cdot \sqrt{4\pi n}}{2\pi n} \]

Simplifying this, we get:

\[ \binom{2n}{n} \approx \frac{4^n \cdot \sqrt{4\pi n}}{2\pi n} = \frac{4^n \cdot 2\sqrt{\pi n}}{2\pi n} = \frac{4^n \cdot \sqrt{\pi n}}{\pi n} = \frac{4^n}{\sqrt{\pi n}} \]

Since \( \sqrt{\pi n} \leq n+1 \) for sufficiently large \( n \), we have:

\[ \frac{4^n}{\sqrt{\pi n}} \geq \frac{4^n}{n+1} \]

Thus, we have shown that:

\[ \binom{2n}{n} \geq \frac{4^n}{n+1} \]

for all \( n \in \mathbb{N} \). This completes the proof.