How can I show that $nb^{n-1}>b^{n-1}+b^{n-2}a+\cdots+a^{n-1}$? I'm trying to show that $$nb^{n-1}>b^{n-1}+b^{n-2}a+\cdots+a^{n-1}$$ when $0<a<b$. I'm aware of $$\begin{align}b>a&\implies a^{n}b^{n-1}\cdot b >a^{n}b^{n-1}\cdot a \\&\implies a^{n}b^{n}>a^{n+1}b^{n-1}\end{align}$$ which suggests that all successive terms following $b^{n-1}$ are less than $b^{n-1}$. However, I don't know how to connect those individual inequalities based on the axioms of ordered field, and its resulting properties to make the final statement.
 A: Take some term and do this :
$$a^ib^{n-1-i}=a \cdot a^{i-1} \cdot b^{n-1-i}<b \cdot a^{i-1} \cdot b^{n-i-1}=a^{i-1} \cdot b^{n-i}$$
Now continue extracting one $a$ at a time to get :
$$a^ib^{n-i-1}<a^{i-1}b^{n-i}<a^{i-2}b^{n-i+1}<\ldots<a \cdot b^{n-2}<b^{n-1}$$
This means every term is less than $b^{n-1}$ and because there are $n$ terms , the whole sum will be less than $nb^{n-1}$ .
EDIT At the request of Il-seob Bae here's a proof of the addition of inequalities on $\mathbb{R}$ .
Suppose $a>b$ and $c>d$ .We need to prove that $a+c>b+d$ .
I'll use three axioms :
$1)$ If $x>y$ then $x+z>y+z$.
$2)$ If $x>y$ and $y>z$ then $x>z$ .
$3)$ $x+y=y+x$ 
From axiom $1)$ :


*

*$a+c>b+c$ because $a>b$ .

*$c+b>d+b$ because $c>d$.


From axiom $3)$ :


*

*$b+c=c+b$ and $d+b=b+d$


Now use the axiom $2)$ of transitivity for $a+c>b+c$ and $b+c>b+d$ to get :
$a+c>b+d$ which proves the claim .
A: Hint : Replace all $a$ s with $b$ s
Since $0<a<b$ ,
$b^{n-1}\cdot a^0+b^{n-2}a^1+b^{n-3}a^2+\cdots+b^0\cdot a^{n-1}<b^{n-1}\cdot b^0+b^{n-2}b^1+b^{n-3}b^2+\cdots+b^0 \cdot b^{n-1}$
$b^{n-1}\cdot a^0+b^{n-2}a^1+b^{n-3}a^2+\cdots+b^0\cdot a^{n-1}<nb^{n-1}$
