<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] 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 $0<p\leq 1/2$
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