Feel free to change the title.
I want to prove that:
$$\mathbb{P}(U_1<U_2<...<U_j, U_j \in(U_{j+1},U_{j+1}+ \beta), U_{j+1}<U_{j+2}<...<U_k) \leq \frac{\beta}{j!(k-j)!}$$
where $U_i$'s are standard uniforms random variables and $\beta$ is a constant between 0 and 1. I was able to prove using stochastically independence conditioning to $U_{j+1}$ \begin{align*} \mathbb{P}(U_1<U_2<...<U_j, U_j \in(U_{j+1},U_{j+1}+ \beta), U_{j+1}<U_{j+2}<...<U_k)\\ = \int_0^1 \mathbb{P}(U_1<U_2<...<U_j, U_j \in(v,v+ \beta), v<U_{j+2}<...<U_k) dv\\ \leq \int_0^1 \mathbb{P}(U_1<U_2<...<U_j, U_j \in(v,v+ \beta))\mathbb{P}( v<U_{j+2}<...<U_k) dv\\ \leq \int_0^1 \mathbb{P}(U_1<U_2<...<U_{j-1})\mathbb{P}( U_j \in(v,v+ \beta))\mathbb{P}( v<U_{j+2}<...<U_k) dv\\ \leq \frac{\beta}{(j-1)!(k-j)!} \end{align*}
But I don't know how can I do to prove the first conjecture.
1. Proof that the desired inequality is false
We write $U_{j+1}=x$, $U_j=y$, $k-j=m$ and $j=n$ with $1\leq m, n$. Let $D_b$, $0<b<1$ be the region $[0,1]^2\cap \{(x,y) \ | \ x<y<x+b\}$. Then the probability mentioned in the title is $$ F(\beta)=\iint_{D_{\beta}} \frac{(1-x)^{m-1}}{(m-1)!} \frac{y^{n-1}}{(n-1)!} \ dA. $$ Let $$G(\beta)=(m-1)!(n-1)!F(\beta)=\iint_{D_{\beta}} (1-x)^{m-1}y^{n-1} \ dA. $$ The wanted inequality is $G(\beta)\leq \beta/(mn)$ while the proved inequality $G(\beta)\leq \beta/m$ is obtained by removing $y^{n-1}$.
By taking the derivative, we have $$ G'(b)=g(b)=\int_{0}^{1-b} (1-x)^{m-1}(x+b)^{n-1} \ dx, $$ and $G(0)=0$. Thus, $G(\beta)=\int_0^{\beta} g(b) \ db$.
Then $g(b)$ is by integration by parts two times when $n\geq 3$, $$ g(b)=\frac{-(1-x)^{m}}m (x+b)^{n-1}\bigg\vert_0^{1-b}-\int_0^{1-b}\frac{-(1-x)^{m}}m(n-1)(x+b)^{n-2} \ dx $$ $$ =\frac{b^{n-1}-b^m}m+\frac{n-1}m\left(\frac{b^{n-2}-b^{m+1}}{m+1}+\int_0^{1-b}\frac{n-2}{m+1}(x-1)^{m+1}(x+b)^{n-3} \ dx\right). $$ However, when $n=2$, the second integration by part stops with $$ g(b)=\frac{b -b^m}m+\frac{1}m \ \frac{1-b^{m+1}}{m+1}. $$ Thus, when $n=2$, we have $$ G(\beta)=\frac{\frac{\beta^2}2-\frac{\beta^{m+1}}{m+1}}{m}+\frac1m \ \frac{\beta-\frac{\beta^{m+2}}{m+2}}{m+1} $$ $$ =\frac{\beta}{2m}\left(\beta+\frac2{m+1}-\frac2{m+1}\beta^m-\frac{2\beta^{m+1}}{(m+1)(m+2)}\right). $$ Taking $\beta=1-\frac 1{m+1}$, we have for sufficiently large $m$, $$ \beta+\frac2{m+1}-\frac2{m+1}\beta^m-\frac{2\beta^{m+1}}{(m+1)(m+2)}= 1+\frac1{m+1}(1-2e^{-1}) + O(\frac1{m^2}). $$ Since $1-2e^{-1}>0$, we have $$\frac1{m+1}(1-2e^{-1}) + O(\frac1{m^2}) >0$$ eventually. Hence, the desired inequality $G(\beta)\leq \beta/(mn)$ is false if $n=2$ and sufficiently large $m$.
2. An expression of the desired probability
Applying the integration by parts repeatedly on the formula for $g(b)$ above, we obtain $$ g(b)=\frac{b^{n-1}-b^m}m+\frac{n-1}m \frac{b^{n-2}-b^{m+1}}{m+1}+\cdots $$ $$ +\frac{n-1}m\frac{n-2}{m+1}\cdots \frac2{m+n-3}\frac{b-b^{m+n-2}}{m+n-2} $$ $$ +\frac{n-1}m\frac{n-2}{m+1}\cdots \frac1{m+n-2}\frac{1-b^{m+n-1}}{m+n-1}. $$ Then term-by-term integration gives $$ F(\beta)=\frac{G(\beta)}{(m-1)!(n-1)!}=\frac1{(m-1)!(n-1)!}\Bigg(\frac{\frac{\beta^n}n-\frac{\beta^{m+1}}{m+1} }{m}+\frac{n-1}m \frac{\frac{\beta^{n-1}}{n-1}-\frac{\beta^{m+2}}{m+2} }{m+1}+\cdots $$ $$ +\frac{n-1}m\frac{n-2}{m+1}\cdots\frac2{m+n-3}\frac{\frac{\beta^2}2-\frac{\beta^{m+n-1}}{m+n-1} }{m+n-2} $$ $$ +\frac{n-1}m\frac{n-2}{m+1}\cdots\frac1{m+n-2}\frac{\beta-\frac{\beta^{m+n}}{m+n}}{m+n-1}\Bigg). $$