<s>[INST] Show that $(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$ is strictly increasing Let $1/2<p< 1$. I am asked to show that
$$f(x)=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}$$
is strictly increasing for $x\geq 0$ and to compute $\lim_{x\to\infty} f(x)$.
I first computed the derivative, but I don't see why it must be positive:
$$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$
Any ideas?
Thanks a lot for your help.

Case $p=3/4$. Then
$$f(x)=(x+1)^{1/4}x^{1/4}(\sqrt{x+1}-\sqrt{x})=\frac{(x+1)^{1/4}x^{1/4}}{\sqrt{x+1}+\sqrt{x}}$$
It suffices to show that $\ln f(x)$ is strictly increasing. We have
$$\ln f(x) =\frac{1}{4} \ln(x+1)+\frac{1}{4} \ln(x)-\ln(\sqrt{x+1}+\sqrt{x})$$
Taking derivative w.r.t. $x$ we get
$$\frac{d \ln f(x) }{d x}=\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x}-\frac{1}{\sqrt{x+1}+\sqrt{x}}(\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}})$$
Hence $\frac{d \ln f(x) }{d x}>0$ is equivalent to
$$(\frac{1}{4}\frac{1}{x+1}+\frac{1}{4}\frac{1}{x})(\sqrt{x+1}+\sqrt{x})>\frac{1}{2}\frac{1}{\sqrt{x+1}}+\frac{1}{2}\frac{1}{\sqrt{x}}$$
which is equivalent to
$$\frac{1}{4}\frac{\sqrt{x}}{x+1}+\frac{1}{4}\frac{\sqrt{x+1}}{x}>\frac{1}{4}\frac{1}{\sqrt{x+1}}+\frac{1}{4}\frac{1}{\sqrt{x}}$$
or
$$\frac{\sqrt{x+1}-\sqrt{x}}{x}>\frac{\sqrt{x+1}-\sqrt{x}}{x+1}$$
which holds. Hence $f(x)$ is strictly increasing.
 [/INST] Rather a long comment than a solution:
My idea was to set $t=p-\frac 12\in(0,\frac 12)$ to introduce symmetry in the formulas, and then rewrite the expression as hyperbolic trig.
$\begin{align}f(x)
&=(x+1)^{\frac 12+t}x^{\frac 12-t}-x^{\frac 12+t}(x+1)^{\frac 12-t}\\\\
&=\sqrt{x}\sqrt{x+1}\,\Big(\tfrac{(x+1)^t}{x^t}-\tfrac{x^t}{(x+1)^t}\Big)\\\\
&=2\sqrt{x}\,\sqrt{x+1}\,\sinh\Big(t\ln(1+\tfrac 1x)\Big)
\end{align}$
We can calculate the derivative and factor out the positive quantity which doesn't change the sign:
$\underbrace{\sqrt{x}\,\sqrt{x+1}\,\cosh\Big(t\ln(1+\tfrac 1x)\Big)}_{\ge 0}\times f'(x)=(2x+1)\tanh\Big(t\ln(1+\tfrac 1x)\Big)-2t$
RHS seems decreasing and positive, but so far I'm stuck at proving it...
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