For $a \gt b \gt 0$ prove that $a^n – b^n \geq n(a-b) (ab)^{\frac{n-1}{2}}$

Question

For $a \gt b \gt 0$ prove that $a^n – b^n \geq n(a-b) (ab)^{\frac{n-1}{2}}$

What is an appropriate way to solve this?

Answer 1

We have $\displaystyle a^n-b^n=(a-b)(\sum_{i=0}^{n-1}a^ib^{n-1-i})$

Now by applying A.M.-G.M on $\displaystyle \sum_{i=0}^{n-1}a^ib^{n-1-i}$ we have,

$\displaystyle \sum_{i=0}^{n-1}a^ib^{n-1-i}\ge n(a^{\sum_{i=0}^{n-1}i}b^{{\sum_{i=0}^{n-1}i}})^{1/n}=n(a^{\frac{(n-1)n}{2}}b^{\frac{(n-1)n}{2}})^{\frac{1}{n}}=n(ab)^{\frac{n-1}{2}}$

So ultimately we have $a^n-b^n\ge n(a-b)(ab)^{\frac{n-1}{2}}$