Prove (a-b)ᵃ⁻ᵇaᵃᵇ⁻ᵃ ≥ (a-1)ᵃᵇ⁻ᵇ

[anemone]

Prove (a-b)ᵃ⁻ᵇaᵃᵇ⁻ᵃ ≥ (a-1)ᵃᵇ⁻ᵇ

Let $a$ and $b$ be positive integers such that $a>b$.

Prove that $(a-b)^{a-b}a^{a(b-1)}\ge (a-1)^{b(a-1)}$.

 

[jvicens]

I understand the importance of providing evidence and logical reasoning to support a claim. In this case, we are trying to prove an inequality involving exponents, so let's start by looking at the properties of exponents.

Firstly, we know that for any positive integers $a$ and $b$, $a^b$ represents the product of $b$ factors of $a$. This means that $a^b$ is always greater than or equal to $a$ itself.

Next, we have the property that $(a^m)^n = a^{mn}$ for any positive integers $a$, $m$, and $n$. This means that we can rewrite our inequality as $(a-b)^{a-b}a^{ab-a} \ge (a-1)^{b^2-b}$.

Now, let's focus on the left side of the inequality. We can rewrite $a-b$ as $(a-1)-(b-1)$, and using the property of exponents mentioned earlier, we have $(a-b)^{a-b} = (a-1)^{a-1}(b-1)^{b-1}$. Similarly, we can rewrite $ab-a$ as $a(b-1)$, so we have $a^{ab-a} = (a^{b-1})^a$. Putting it all together, our left side becomes $(a-1)^{a-1}(b-1)^{b-1}(a^{b-1})^a$.

Now, let's focus on the right side of the inequality. Using the property of exponents again, we can rewrite $a-1$ as $(a-1)^{b-1}(a-1)$, and $b^2-b$ as $b(b-1)$. Putting it all together, our right side becomes $(a-1)^{b^2-b} = (a-1)^{b(b-1)}(a-1)^{b-1}$.

Finally, we can see that the left side of our inequality is a product of three terms, while the right side is a product of only two terms. From our earlier discussion about the properties of exponents, we know that a product of more terms will always be greater than or equal to a product of fewer terms with the same base. Therefore, we can conclude that $(a-1)^