If $a > b$ and $x > 2$ show $\frac{2^{a+1}-1}{2^a}\frac{x^{b+1}-1}{x^b(x-1)} > \frac{2^{b+1}-1}{2^b}\frac{x^{a+1}-1}{x^a(x-1)}$ Given integers $a > b \geq 0$ and $p>2$, show that the following inequality holds
$$\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b(x-1)} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a(x-1)}$$
This is trivial if $b=0$ as we'll end up with the inequality 
$$\dfrac{2^{a+1}-1}{2^a} > \dfrac{x^{a+1}-1}{x^a(x-1)}$$
which can be shown using a variety of methods such as taking the derivative of both sides. I want to show that this inequality holds for all values of $a,b,x$ which I am able to plot out trivially, however I get stuck down the line. 
Here is what I've tried so far:
$$\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b(x-1)} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a(x-1)}\equiv\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a}$$
then we cross multiply to find
$$
\dfrac{2^b}{2^{b+1}-1}\dfrac{2^{a+1}-1}{2^a} > \dfrac{x^b}{x^{b+1}-1}\dfrac{x^{a+1}-1}{x^a} \equiv \dfrac{2^{a+b+1}-2^b}{2^{a+b+1}-2^a}>\dfrac{x^{a+b+1}-x^b}{x^{a+b+1}-x^a}
$$
but from here I find myself stuck without a direction forward. Any help would be appreciated!

$
f(x)=\dfrac{x^{a+b+1}-x^b}{x^{a+b+1}-x^a}=\dfrac{1-\frac{1}{x^{a+1}}}{1-\frac{1}{x^{b+1}}}\\
f'(x)=\dfrac{\frac{a+1}{x^{a+2}}-\frac{b+1}{x^{b+2}}-\frac{a-b}{x^{a+b+3}}}{\left(1-\frac{1}{x^{b+1}}\right)^2}=\dfrac{(a+1)x^{b+1}-(b+1)x^{a+1}-a+b}{x^{a+b+3}\left(1-\frac{1}{x^{b+1}}\right)^2}\\
g(x)=(a+1)x^{b+1}-(b+1)x^{a+1}-a+b\\
g(1)=0\\g'(x)=(a+1)(b+1)(x^b-x^a)\\
x>1\,\Rightarrow\,g'(x)<0\,\Rightarrow\,g(x)<0\,\Rightarrow\,f'(x)<0\,\Rightarrow\,f(2)>f(x)
$
