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.
 A: Computing the limit is rather straight forward
\begin{align}
f(x)&=(x+1)^{p}x^{1-p}-(x+1)^{1-p}x^{p}=\frac{(x+1)^p}{x^p}x-(x+1)\frac{x^p}{(1+x)^p}\\
&=\frac{(1+x)^{2p}x-(x+1)x^{2p}}{x^p(1+x)^p}=\frac{x\Big(\big(1+\tfrac1x\big)^{2p} -1\Big)-1}{\big(1+\frac{1}{x}\big)^p}\\
&=\frac{\frac{\big(1+\tfrac1x\big)^{2p}-1}{\tfrac1x} -1}{\big(1+\tfrac1x\big)^p}\xrightarrow{x\rightarrow\infty}2p-1
\end{align}
As for monotonicity, it is easier to set $t=1/x$ to obtain that $f(x)=\phi(1/x)$ where
$$\phi(t)=\frac{(1+t)^{2p}-1-t}{t(1+t)^p}=\frac{(1+t)^{2p-1}-1}{t(1+t)^{p-1}}=\frac{(1+t)^p-(1+t)^{1-p}}{t}$$
for $t>0$. Notice that $\phi(t)>0$ for all $t>0$ and so,  $\xi(t)=(1+t)^p-(1+t)^{1-p}>0$ for all $t>0$.
The function $\xi$ is concave (i.e. $-\xi$ is convex) since
\begin{align}
\xi''(t)&=p(p-1)(1+t)^{p-2}+p(1-p)(1+t)^{-p-1}\\
&=\frac{p(1-p)(1+t)^{-1}}{(1+t)^p(1+t)^{1-p}}\Big((1+t)^{1-p}-(1+t)^{-p}\Big)\\
&=-\frac{p(1-p)(1+t)^{-1}}{(1+t)^p(1+t)^{1-p}}\,\xi(t)<0
\end{align}
Consequently, the map $t\mapsto \frac{\xi(t)-\xi(0)}{t}=\phi(t)$ is monotone decreasing. Hence, $f(x)=\phi(1/x)$ is monotone increasing.
This is based on a well known fact that if $g$ is a convex function on an interval $I$ and $t_0\in I$, then $t\mapsto \frac{g(t)-g(t_0)}{t-t_0}$ is increasing  on $(t_0,\infty)$.
A: 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...
A: As zwim just a comment :
Starting with your derivative and using the lemma 7.1 (see the reference) we have :
$$\frac{px}{x+1}\left(\left(1-p\right)^{2}+\frac{x+1}{x}p(2-p)-\frac{x+1}{x}p(1-p)\ln\frac{x+1}{x}\right)\leq p(x+1)^{p-1}x^{1-p}$$
Or :
$$\frac{p(1-p)^{2}x}{x+1}+p^{2}(2-p)-p^{2}(1-p)\ln\frac{x+1}{x}\le p(x+1)^{p-1}x^{1-p}$$
In the lemma we have : $$a=\frac{x+1}{x},c=p$$
We can do it for the four partial expression .
I think there is some simplifications at the end .
Reference :
VASILE CIRTOAJE, PROOFS OF THREE OPEN INEQUALITIES WITH POWER-EXPONENTIAL FUNCTIONS, Journal of Nonlinear Sciences and Applications, 4 (2011), no. 2, 130-137


Another way :
Considering $a>0$ fixed and the function for $0.5\leq x\leq 1$ :
$$g(x)=(a+1)^{x-1}a^{1-x}-(a+1)^{x}a^{-x}+(a+1)^{-x}a^{x}-(a+1)^{1-x}a^{x-1}$$
$$g'(x)=-a^{\left(-x-1\right)}(a+1)^{\left(x-1\right)}(a^{\left(2x+1\right)}+a^{2x}-a(a+1)^{\left(2x\right)})(\log(a)-\log(a+1))\leq 0$$
So the function in $x$ is decreasing so $$f(1)\leq f(x)$$
Now consider your derivative :
$$\frac{d f(x)}{d x}=p(x+1)^{p-1}x^{1-p}+(1-p)(x+1)^px^{-p}-(1-p)(x+1)^{-p}x^p -p(x+1)^{1-p}x^{p-1}$$
We have :
$$\frac{d f(x)}{d x}=p\left((x+1)^{p-1}x^{1-p}-(x+1)^{p}x^{-p}\right)+(x+1)^{p}x^{-p}+p\left((x+1)^{-p}x^{p}-(x+1)^{1-p}x^{p-1}\right)-(x+1)^{-p}x^{p}$$
Or :
$$\frac{d f(x)}{d x}=p\left((x+1)^{p-1}x^{1-p}-(x+1)^{p}x^{-p}+(x+1)^{-p}x^{p}-(x+1)^{1-p}x^{p-1}\right)+(x+1)^{p}x^{-p}-(x+1)^{-p}x^{p}$$
Now use $g(x)$ and another function like $g(x)$ to conclude partially.
In fact I show that $f'(x)\leq 0$ for $0<p\leq 1/2$
