If $a < b$ then $\sum_{k = 0}^{2n - 2} a^k b^{2n - 2 - k} \geq 0$

Question

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?

Answer 1

If $a=0$, so our inequality is obviously true.

For $a\neq0$ we obtain $$\sum_{k=0}^{2n-2}a^kb^{2n-2-k}=a^{2n-2}(x^{2n-2}+x^{2n-3}+...+x+1),$$ where $x=\frac{b}{a}.$

Since $a^{2n-2}>0$, it's enough to prove that: $$x^{2n-2}+x^{2n-3}+...+x+1\geq0.$$ Now, consider three cases:

1. $x>0$;

2. $-1\leq x\leq0$;

3. $x<-1$.

Can you end it now?

Answer 2

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.