If $0<\beta<1$ and $0<x<1,$ how to prove that $$h(x)-2x+(4-2^{1+\beta})x^{1+\beta}<0,$$ where $$h(x)=(1+x)^{1+\beta}-x^{1+\beta}-1.$$The numerical simulation shows that it is true.



A proof of an interesting inequality
$\newcommand\be\beta$We have to show that
$$g(x):=(1+x)^{1+\be}+(3-2^{1+\be}) x^{1+\be}-2 x-1<0; \tag{1}\label{1}$$
here and in what follows, $\be$ and $x$ are in $(0,1)$.


Let $r(x):=\dfrac x{1+x}$, so that $r$ is increasing on $(0,1)$.
Hence, $$g''(x)=(1+\be)\be x^{\be-1}(r(x)^{1-\be}+3-2^{1+\be})$$
can only switch sign from $-$ to $+$ (if at all) as $x$ increases from $0$ to $1$.
Noting also that $g(0)=g(1)=0$ and $g'(0)=\be-1<0$, we get \eqref{1}.


