I want to prove the monotonicity of $I_r(nr, 2+(1-r)n)$ on $n$ but has no clues. The $I$ is the regularized incomplete beta function, defined as follows: $$I_r(nr, 2+(1-r)n)=\frac{\int_0^r x^{nr-1}(1-x)^{1+(1-r)n}dx}{\int_0^1 x^{nr-1}(1-x)^{1+(1-r)n}dx}$$ where $n>0$ and $r\in[0,1]$.
It seems that $\frac{\partial}{\partial n}I_r(nr, 2+(1-r)n)<0$ based on Mathematica.
Denote $I_0(n)=\int_0^r x^{nr-1}(1-x)^{1+(1-r)n}dx$, $I_1(n)=\int_r^1 x^{nr-1}(1-x)^{1+(1-r)n}dx$. You want to prove that $I_0/(I_0+I_1)$ decreases, equivalently, $(I_0+I_1)/I_0$ increases, equivalently, $I_1/I_0$ increases.
Denote $x^r(1-x)^{1-r}=:h(x)$. Then $$h'(x)=h(x)\cdot \left(\frac rx - \frac{1-r}{1-x}\right)=h(x)\cdot \frac{r-x}{x(1-x)}.$$ Thus, $h$ increases on $[0,r]$ from 0 to $t_0:=h(r)$ and decreases on $[r,1]$ from $t_0$ to 0. Therefore, in both integrals $I_0(n)$, $I_1(n)$ we may use the change of variables $t=h(x)$, and $t$ varies from 0 to $t_0$. Denote $x=f(t)$ for $I_0$ and $x=g(t)$ for $I_1$. Then \begin{align} I_0(n)&=\int_0^{t_0} t^n\frac{1-f(t)}{f(t)}f'(t)dt&=:\int_0^{t_0} t^nA(t)dt,\\ I_1(n)&=\int_0^{t_0} t^n\frac{1-g(t)}{g(t)}(-g'(t))dt&=:\int_0^{t_0} t^nB(t)dt .\end{align} In order to prove that $I_1/I_0$ increases, we apply the following
Lemma. If $A(t)$, $B(t)$ are positive (measurable) functions on $(0,t_0)$ and $B/A$ increases on $(0,t_0)$, then $I_1(n)/I_0(n)$ increases as a function of $n>0$, where $I_1$, $I_0$ are defined as above integrals.
Proof. We should prove that $I_1(n)/I_0(n)\leqslant I_1(n+m)/I_0(n+m)$ for all $n,m>0$. Equivalently, we want $I_1(n)I_0(m+n)\leqslant I_1(n+m)I_0(n)$. Well, $$I_1(n)I_0(m+n)=\int_0^{t_0}t^nB(t)dt\int_0^{t_0}s^{m+n}A(s)ds \\=\int_{0<t<s<t_0} t^ns^{m+n}A(s)B(t)+s^nt^{m+n}A(t)B(s)dsdt.$$ Write in analogous fashion $I_1(n+m)I_0(n)$ and note that for all $t<s$ we have $$ t^ns^{m+n}A(t)B(s)+s^nt^{m+n}A(s)B(t)-t^ns^{m+n}A(s)B(t)-s^nt^{m+n}A(t)B(s) =t^ns^n(s^m-t^m)(A(t)B(s)-A(s)B(t))\geqslant 0, $$ since $B/A$ being increasing yields $A(t)B(s)\geqslant A(s)B(t)$. So, $\Phi:=I_1(n+m)I_0(n)-I_1(n)I_0(n+m)$ is an integral of a non-negative expression over the triangle $0<s<t<t_0$, and $\Phi\geqslant 0$ as needed.
It remains to prove that $B/A$ increases on $(0,t_0)$. From $h(f(t))=t$ we get $$1=f'(t)h'(f(t))=f'(t)h(f(t))\cdot\frac{h'(f(t))}{h(f(t))} =tf'(t)\cdot\frac{r-f(t)}{f(t)(1-f(t))}.$$ Therefore $A(t)=\frac{(1-f(t))^2}{t(r-f(t))}$, analogously $B(t)=\frac{(1-g(t))^2}{t(g(t)-r)}.$ We need to prove that $B/A$ increases, equivalently, $(\log(B/A))'\geqslant 0$. This reads as $$ -2\frac{g'(t)}{1-g(t)}-\frac{g'(t)}{g(t)-r}+2\frac{f'(t)}{1-f(t)}-\frac{f'(t)}{r-f(t)}\geqslant 0 \Leftrightarrow\\ -g'(t)\frac{(g(t)+1-2r)}{(1-g(t))(g(t)-r)}\geqslant f'(t)\frac{(f(t)+1-2r)}{(1-f(t))(r-f(t))}\Leftrightarrow\\ \frac{g(t)(g(t)+1-2r)}{(g(t)-r)^2}\geqslant \frac{f(t)(f(t)+1-2r)}{(r-f(t))^2} \Leftrightarrow\\ g(t)(g(t)+1-2r)(r-f(t))^2 \geqslant f(t)(f(t)+1-2r)(g(t)-r)^2. $$ Ok, we have $$(r-f(t))^2=\int_t^{t_0}2(r-f(s))f'(s)ds=2\int_t^{t_0}\frac{f(s)(1-f(s))}sds,$$ analogously for $g$. Thus, it suffices to prove that for every $s\in (t,t_0)$ we have $$ g(t)(g(t)+1-2r)f(s)(1-f(s))\geqslant f(t)(f(t)+1-2r)g(s)(1-g(s)).\tag{1} $$ Note that $g(t)\geqslant g(s)\geqslant r$. The quadratic trinomial $\eta(x):=x(x+1-2r)$ increases on $[r-1/2,\infty)$, thus $g(t)(g(t)+1-2r)= \eta(g(t)) \geqslant \eta(g(s))=g(s)(g(s)+1-2r)$. Also, $f(t)\in [0,f(s)]$, thus $\eta(f(t))\leqslant \max(\eta(0),\eta(f(s)))=\max(0,\eta(f(s)))$. Since LHS of (1) is non-negative, it remains to prove that $$ g(s)(g(s)+1-2r)f(s)(1-f(s))\geqslant f(s)(f(s)+1-2r)g(s)(1-g(s)). \tag{2} $$ Since $g(s)\geqslant f(s)$, we have $g(s)+1-2r\geqslant f(s)+1-2r$, $1-f(s)\geqslant 1-g(s)$, also $g(s)+1-2r\geqslant 0$, and (2) follows.
