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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 $b > a > 0$ 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?

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$[1]$ 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 $f'(x) > 0$, $g'(x) > 0 $ for all $x > 0$. We need to show ($\frac{f(x)}{g(x)})' > 0 $ for all $x$. but $$(\frac{f(x)}{g(x)})' = \frac{f'g - fg'}{g^2} > 0 \iff f'g > g'f $$

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

$$\therefore f'g > g'f \iff \frac{f'}{g'} > \frac{f}{g} > 0 $$

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  • $\begingroup$ I'm pretty sure [5] is the whole problem. $\endgroup$ – Karolis Juodelė Nov 1 '13 at 9:22
  • $\begingroup$ ok how about now $\endgroup$ – ILoveMath Nov 1 '13 at 9:25
  • $\begingroup$ $(\frac f g)' \neq \frac {f'} {g'}$. Consider $f = x$ and $g = x^2$. $\endgroup$ – Karolis Juodelė Nov 1 '13 at 9:37
  • $\begingroup$ You have $$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}$$ $\endgroup$ – Daniel Fischer Nov 1 '13 at 9:37
  • $\begingroup$ fixed it now.... $\endgroup$ – ILoveMath Nov 1 '13 at 9:58

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