<s>[INST] Prove that $\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$ is strictly increasing The problem is to prove that
$$\left(0.5^x-0.25^x\right)^{\frac{1}{x}}$$
is strictly increasing on $\mathbb{R}^+$. The problem I'm having is that once I found the derivative to be
$$\left(\dfrac{1}{2^x}-\dfrac{1}{4^x}\right)^\frac{1}{x}\left(\dfrac{\frac{\ln\left(4\right)}{4^x}-\frac{\ln\left(2\right)}{2^x}}{x\left(\frac{1}{2^x}-\frac{1}{4^x}\right)}-\dfrac{\ln\left(\frac{1}{2^x}-\frac{1}{4^x}\right)}{x^2}\right),$$
I don't know how to show that this expression is always positive. Any help?
 [/INST] Assume $a$ is a real number such that $a>1$. Then let
$$
f_a(x):=\left(1-\frac1{a^x}\right)^{1/x}, \qquad x>0.
$$ We have
$$
\begin{align}
f_a'(x)&=\left( -\frac1{x^2}\cdot \ln \left(1-\frac1{a^x}\right)+\frac1x \cdot \frac{\ln a \cdot \frac1{a^x}}{1-\frac1{a^x}} \right)\cdot\left(1-\frac1{a^x}\right)^{1/x}
\\\\&=\frac1x \left( -\frac1{x}\cdot \ln \left(1-\frac1{a^x}\right)+ \frac{\ln a }{a^x-1} \right)\cdot\left(1-\frac1{a^x}\right)^{1/x}
\\\\&>0 \quad (\text{a sum of positive terms})
\end{align}
$$ thus $f_a$ is increasing over $(0,\infty)$. Now write
$$
\left(0.5^x-0.25^x\right)^{\frac{1}{x}}=\frac12 \cdot\left(1-\frac1{2^x} \right)^{1/x}=\frac12 f_2(x)
$$ to conclude.
</s>