Increasing function by derivative? I would like to show the following function is increasing for $x \geq 1$
$$
x \frac{(b+x)^k - (a+x)^k}{(b+x)^k - a^k}
$$
where $b &gt; a &gt; 0$ and integer $k \geq 1$. It is easy to prove by derivative for $k=2,3$ as we can write explicitly the derivative. However, it is not clear for general $k$. Do you have any idea?
$[1]$ If $f$ and $g$ are increasing functions, then $f+g$ and $fg$ $f + Constant$ are increasing. Check yourself this.

$[2]$: $f(x) = x $ in increasing. Obvious.

$[3]$: for $x \geq 1$ ,$f(x) = (b +x)^k$ is increasing for positive $k$. To see this, $f'(x) = k(b+x)^{k-1} \geq 0 $ 

$[4]$: $f(x) = (a + x)^k$ is also increasing as in $[3]$

$[5]$ is you can show that $\frac{f}{g}$ is increasing given $f$ and $g$ increasing, then the problem is solved.

To see $[5]$ suppose $f'(x) &gt; 0$, $g'(x) &gt; 0 $ for all $x &gt; 0$. We need to show ($\frac{f(x)}{g(x)})' &gt; 0 $ for all $x$. 
but $$(\frac{f(x)}{g(x)})' = \frac{f'g - fg'}{g^2} &gt; 0 \iff f'g &gt; g'f $$

So we need to check $ f'g &gt; g'f$. We can assume in your problem $f,g &gt; 0$ since $b &gt; a&gt;0$

$$\therefore f'g &gt; g'f \iff \frac{f'}{g'} &gt; \frac{f}{g} &gt; 0 $$
$[1]$ If $f$ and $g$ are increasing functions, then $f+g$ and $fg$ $f + Constant$ are increasing. Check yourself this.

$[2]$: $f(x) = x $ in increasing. Obvious.

$[3]$: for $x \geq 1$ ,$f(x) = (b +x)^k$ is increasing for positive $k$. To see this, $f'(x) = k(b+x)^{k-1} \geq 0 $ 

$[4]$: $f(x) = (a + x)^k$ is also increasing as in $[3]$

$[5]$ is you can show that $\frac{f}{g}$ is increasing given $f$ and $g$ increasing, then the problem is solved.

To see $[5]$ suppose $f'(x) &gt; 0$, $g'(x) &gt; 0 $ for all $x &gt; 0$. We need to show ($\frac{f(x)}{g(x)})' &gt; 0 $ for all $x$. 
but $$(\frac{f(x)}{g(x)})' = \frac{f'g - fg'}{g^2} &gt; 0 \iff f'g &gt; g'f $$

So we need to check $ f'g &gt; g'f$. We can assume in your problem $f,g &gt; 0$ since $b &gt; a&gt;0$

$$\therefore f'g &gt; g'f \iff \frac{f'}{g'} &gt; \frac{f}{g} &gt; 0 $$