I'm trying to prove that:
$$
\left(1-\frac{1}{x}\right)^{\lceil x\ln\left(2\right)\rceil}-\left(1-\frac{2}{x}\right)^{\lceil x\ln\left(2\right)\rceil} \geq \frac{1}{4}
$$
for every $x\geq 1$.
If the assumptions $x\geq 2$ and/or $x\in\mathbb{N}$ make the proof easier, I can use them.
My attempt:
I know that:
$$
\left(1-\frac{1}{x}\right)^{x\ln\left(2\right)}-\left(1-\frac{2}{x}\right)^{x\ln\left(2\right)} \rightarrow_{x\rightarrow \infty} \frac{1}{4}
$$
and that this happens monotonically from above ($\geq\frac{1}{4}$).
But I'm not sure how I can do it with the ceiling function.
Let $f_k(x) = \left(1-\dfrac{1}{x}\right)^k - \left(1-\dfrac{2}{x}\right)^k$ for fixed $k\in\mathbb{Z}$, then $f'_k(x) = \dfrac{k}{x^2}\left(\left(1-\dfrac{1}{x}\right)^{k-1} - 2\left(1-\dfrac{2}{x}\right)^{k-1}\right)$. Suppose that $k\ge 2$. For even $k$, $f'_k(x)=0$ has only one solution $x_k$ such that $\dfrac{x_k-1}{x_k-2}=\sqrt[k-1]{2}$, or $x_k=2+\dfrac{1}{\sqrt[k-1]{2}-1}$. It is easy to see that $f_k$ increases over $(0,x_k)$ and decreases over $(x_k,+\infty)$. For odd $k$, $f'_k=0$ has two solutions $x_k$ and $x'_k$ such that $\dfrac{x_k-1}{x_k-2}=\sqrt[k-1]{2}$ and that $\dfrac{x_k-1}{x_k-2}=-\sqrt[k-1]{2}$, or $x_k=2+\dfrac{1}{\sqrt[k-1]{2}-1}$ and $x'_k=2-\dfrac{1}{\sqrt[k-1]{2}+1}$. $f_k$ decreases over $(0,x'_k)$, increases over $(x'_k,x_k)$, and decreases over $(x_k,+\infty)$.
Now your $f(x)$ equals $f_k(x)$ over $\left(\dfrac{k-1}{\ln 2},\dfrac{k-1}{\ln 2}\right]$. The point is that $x_k>\dfrac{k}{\ln 2}$ for $k\ge 2$ (see ($\star$) below), so we have $\left(\dfrac{k-1}{\ln 2},\dfrac{k-1}{\ln 2}\right]\subset (0,x_k)$ for even $k$ and $\left(\dfrac{k-1}{\ln 2},\dfrac{k-1}{\ln 2}\right]\subset (x'_k,x_k)$ for odd $k$ (since $\dfrac{k-1}{\ln 2}\ge \dfrac{2}{\ln 2}>2>x'_k$). As a result, $f(x)$ is decreasing over $\left(0,\dfrac{1}{\ln 2}\right]$, and is increasing over $\left(\dfrac{k-1}{\ln 2},\dfrac{k-1}{\ln 2}\right]$ for every $k\ge 2$.
For $k=2$, we have $f(x) = \left(1-\dfrac{1}{x}\right)^2 - \left(1-\dfrac{2}{x}\right)^2 = \dfrac{2}{x}-\dfrac{3}{x^2}$ over $\left(\dfrac{1}{\ln 2},\dfrac{2}{\ln 2}\right]$. It is easy to see that $f(x)\ge\dfrac{1}{4}$ if and only if $x\in\left[2,\dfrac{2}{\ln 2}\right]$. For $k\ge 3$, we have $$\inf_{x\in\left(\frac{k-1}{\ln 2},\frac{k-1}{\ln 2}\right]}f(x) = f_k\left(\dfrac{k-1}{\ln 2}\right) = \left(1-\dfrac{\ln 2}{k-1}\right)^k - \left(1-\dfrac{2\ln 2}{k-1}\right)^k,$$ which can shown to be greater than $\dfrac{1}{4}$ (see ($\star\star$) below$).
$\star$: Write $t=\dfrac{\ln 2}{k-1}$, then we need to show $2+\dfrac{1}{\mathrm{e}^t-1}>\dfrac{1}{t}+\dfrac{1}{\ln 2}$ for $t\in(0,\ln 2)$. Note that $$\left(2+\dfrac{1}{\mathrm{e}^t-1}\right)-\left(\dfrac{1}{t}+\dfrac{1}{\ln 2}\right) = \dfrac{3}{2}-\dfrac{2}{\ln 2} +\left(\operatorname{coth}x-\dfrac{1}{x}\right).$$ It is easy to see that $\operatorname{coth}x>\dfrac{1}{x}$ for $x>0$, and that $\ln 2<\dfrac{3}{4}$ (by $\ln x<\dfrac{1}{2}\left(x-\dfrac{1}{x}\right)$ for $x>1$).
$\star\star$: Write $t=\dfrac{\ln 2}{k-1}$, then we need to show $(1-t)^{1+\frac{\ln 2}{t}}-(1-2t)^{1+\frac{\ln 2}{t}}\ge\dfrac{1}{4}$ for $t=\dfrac{\ln 2}{2},\dfrac{\ln 2}{3},\cdots$.
Claim 1:$(1-2t)^{\frac{\ln 2}{t}}\le\dfrac{1}{4}-\dfrac{\ln 2}{2}t$, or $\ln(1-2t)\le\dfrac{t}{\ln 2}\ln\left(\dfrac{1}{4}-\dfrac{\ln 2}{2}t\right)$. Write $p(t) = \dfrac{t}{\ln 2}\ln\left(\dfrac{1}{4}-\dfrac{\ln 2}{2}t\right)-\ln(1-2t)$, then $$p''(t) = \dfrac{x\left(x-\frac{1}{2}\right)\left(x-\left(-\ln 2+\frac{1}{2}+\frac{1}{\ln 2}\right)\right)+\frac{x}{2}(1-\ln 2)^2}{(1-2x)^2\left(\frac{1}{4}-\frac{\ln 2}{2}x\right)^2}.$$ We can see $p''(t)>0$ for $t\in \left(0,\dfrac{1}{2}\right)$, so $p'(t)>p'(0)=0$, and $p(t)>p(0)=0$ for $t\in \left(0,\dfrac{1}{2}\right)$.
Claim 2:$(1-t)^{1+\frac{\ln 2}{t}}\ge\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{\ln 2}{4}\right)t$, or $(t+\ln 2)\ln(1-t)\ge t\ln\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{\ln 2}{4}\right)t$. Write $q(t) = t\ln\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{\ln 2}{4}\right)-(t+\ln 2)\ln(1-t)$, then $$q''(t) = \dfrac{\frac{x\ln 2}{16}\left((\ln^2 2+4\ln 2+4)x-(3\ln 2+4)\right)}{(1-t)^2\left(\dfrac{1}{2}+\dfrac{\ln 2}{4}\right)^2}.$$ Since $\dfrac{3\ln 2+4}{\ln^2 2+4\ln 2+4}>\dfrac{1}{2}$, we can see $q''(t)<0$ for $t\in \left(0,\dfrac{1}{2}\right)$, so $q'(t)<q'(0)=0$, and $q(t)<q(0)=0$ for $t\in \left(0,\dfrac{1}{2}\right)$.
Now $\dfrac{1}{2}-\left(\dfrac{1}{2}+\dfrac{\ln 2}{4}\right)t\ge (1-2t)\left(\dfrac{1}{4}-\dfrac{\ln 2}{2}t\right)+\dfrac{1}{4}$ for $t\in \left[0,\dfrac{1}{4}\right]$, so ($\star\star$) is proved for $\dfrac{\ln 2}{k-1}\le\dfrac{1}{4}$, or $k\ge 4$. It remains to consider the case $k=3$, or $\left(1-\dfrac{\ln 2}{2}\right)^3 - \left(1-\dfrac{2\ln 2}{2}\right)^3\ge\dfrac{1}{4}$. This is equivalent to prove that $g(\ln 2)>0$ for $g(t) = 7t^3-18t^2+12t-2$. Note thst $g(t)$ is decreasing around its root near $\ln 2$. We have $$\mathrm{e}^{\frac{52}{75}}>1+\dfrac{52}{75}+\dfrac{1}{2!}\left(\frac{52}{75}\right)^2+\dfrac{1}{3!}\left(\frac{52}{75}\right)^3+\dfrac{1}{4!}\left(\frac{52}{75}\right)^4+\dfrac{1}{5!}\left(\frac{52}{75}\right)^5=\dfrac{71198573129}{35595703125}>2,$$ or $\ln 2<\dfrac{52}{75}$, so $$g(\ln 2)>g\left(\frac{52}{75}\right)=\dfrac{106}{421875}>0.$$