Proving the Inequality 2n 2! ≥ n 2! 2^n

Proving mathematical inequalities is a fundamental part of mathematics, particularly in areas such as combinatorics and discrete mathematics. One such inequality is 2n 2! ≥ n 2! 2^n. In this article, we will delve into the detailed proof of this inequality by breaking down each step and understanding the underlying principles.

Step 1: Rewrite the Inequality

The given inequality can be rewritten as:

$$frac{2n 2!}{n 2!} geq n 2^n$

Our goal is to show that the left-hand side is always greater than or equal to the right-hand side.

Step 2: Simplify the Left-Hand Side

By using the definition of factorials, we can simplify the left-hand side as follows:

$$frac{2n 2!}{n 2!} 2n 2 2n - 1 cdot 2n - 2 cdot cdots cdot n 3$

This expression consists of n 2 terms starting from 2n 2 down to n 3.

Step 3: Analyze the Right-Hand Side

The right-hand side of the inequality is n 2^n, which is an exponential function.

Step 4: Compare Terms

Now, we need to demonstrate that the product of the left-hand side is greater than or equal to n 2^n. For each k from 0 to n-1:

$$2n 2 - k geq n 2$

Let's analyze how many terms we have:

From 2n 2 to n 3, there are n 2 terms.

Step 5: Establish the Inequality

It is beneficial to show that each term in the product on the left-hand side is at least n 2 for a sufficiently large n:

The smallest term in the product is when k n, resulting in 2n 2 - n - 2 n.

Each term from 2n 2 down to n 3 is greater than or equal to n 2 for large n.

Step 6: Conclusion

Based on the analysis, we can conclude that:

$$2n 2! geq n 2! 2^n$

This holds true because the left-hand side grows significantly faster than the right-hand side as n increases due to the factorials rapid growth. Therefore, the inequality 2n 2! ≥ n 2! 2^n is indeed true for all n ≥ 0.

Understanding the growth rates of factorials and exponentials is crucial for tackling inequalities in combinatorial proofs and discrete mathematics.