How to show $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$? I am trying to show that the function $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$, where $0.5 \leq L < H < 1$. Visual inspection in Mathematica suggests the conjecture is true.  I tried showing this by differentiation, but am having trouble signing it.  How do I show that $\frac{(1-L)^x - (1-H)^x}{H^x-L^x}$ is decreasing in $x$?  Thanks!
 A: You can write
$$
f(x) = \frac{(1-H)^x - (1-L)^x}{H^x - L^x} = \left(\frac{1-L}{H}\right)^x \cdot 
\frac{1 - \left(\frac{1-H}{1-L}\right)^x}{1 - \left(\frac{L}{H}\right)^x}=
u(x)\cdot v(x).
$$
Then use the inequalities to obtain,  $$(1-L)/H < 1 \quad\text{and}\quad
0 < \frac{1-H}{1-L}<\frac{L}{H}<1.$$
These inequalities imply both $u(x), v(x)$ are positive;  moreover, from the first $u(x)$ is plainly decreasing and, with a little further investigation, the second means $v(x)$ is also decreasing.  we can therefore conclude $f(x)$ is decreasing.
It remains to show $v(x)$ is decreasing.
We remark that $\log t / (t-1)$ is a decreasing function of $t$ for $ t > 0$, because its derivative is,
$$
\frac{\log(1/t) - 1/t +1 }{(t-1)^2} 
$$
which is strictly negative when $t>0$.
Now to verify $v(x)$ is decreasing, introduce $a= (1-H)/(1-L), b = L/H$ so that $0 < a < b < 1$.  Then, for $x >0$, the derivative,
\begin{align}
v'(x) &= \frac{-\log(a)a^x (1-b^x) + \log(b) b^x(1-a^x)}{(1-b^x)^2} \\
\end{align}
Focus on the numerator, and consider the ratio
\begin{align}
\frac{-\log(a) a^x (1-b^x)}{-\log(b) b^x (1-a^x)} &= \frac{x\log(1/a)(1/b^x - 1)}{x\log(1/b)(1/a^x-1)} \\
&=\frac{\log(1/a^x)}{1/a^x - 1} \cdot \frac{1/b^x - 1}{\log(1/b^x)}
\end{align}
As $x \to 0$, this expression has limit $1$ since left and right terms converge to the derivative of $\log(t) $ at $t=1$ and its reciprocal respectively.  But we also have $1/b^x < 1/a^x$ when $x > 0$ and using our remark above,
\begin{align*}
  \frac{\log(1/a^x)}{1/a^x - 1} < \frac{\log(1/b^x)}{1/b^x - 1}
 \end{align*}
so that
\begin{align}
\frac{-\log(a) a^x (1-b^x)}{-\log(b) b^x (1-a^x)} < 1
\end{align}
from which it follows $v'(x) < 0$ when $x > 0$.
That leaves uncertainty when $x=0$ at which point $f$ and $v$ are not properly defined.  However, using the power series for $e^z$, we can see
$$
v(x) = \frac{\log\left(\tfrac{1-H}{1-L}\right)x + \tfrac{1}{2!} \log\left(\tfrac{1-H}{1-L}\right)^2x^2 + O(x^3)}{\log\left(\tfrac{L}{H}\right)x + \tfrac{1}{2!} \log\left(\tfrac{L}{H}\right)^2 x^2 + O(x^3)}.$$
Because $L \neq H$ it is plain $v(x)$ thus extends to a continuously differentiable function at $x=0$, so that the derivative at $x=0$ would also be negative or zero.
Accordingly $v(x)$ is decreasing for $x \geqslant 0$.
