If $a < b$ then $\sum_{k = 0}^{2n - 2} a^k b^{2n - 2 - k} \geq 0$ Recently I've been thinking how to prove that $f(x) = x^{2n - 1}$ is monotonic increasing for all positive integer $n$ by directly using the definition. I found that the following inequality must hold: $$a < b \implies\sum_{k = 0}^{2n - 2} a^k b^{2n - 2 - k} \geq 0.$$
If I can prove this, the problem is also solved. Any hint?

For $a<b$,
$$\sum_{k = 0}^{2n - 2} a^k b^{2n - 2 - k}=\frac{b^{2n-1}-a^{2n-1}}{b-a}.$$
Both the numerator and denominator of this fraction are positive.
