Resolution of an inequality on integers I’m trying to resolve respect to $k$ the following inequality, 
$$
k\left(\log k +\log \log k-\alpha+O\left(\frac{\log \log k}{\log k}\right)\right)\geq x,
$$ 
in order to obtain, under the condition $k\leq x$, that
$$
\ k\geq \dfrac{x}{\log x}\left(1+\dfrac{\alpha+o(1)}{\log x}\right)
$$ 
is this possible? 
 A: $\newcommand\ka{\kappa}$
Let $a:=\alpha$. Let $f$ be a function such that 
$$f(k)=k\Big(\ln k +\ln\ln k-a+O\Big(\frac{\ln\ln k}{\ln k}\Big)\Big)
=k\big(\ln k +\ln\ln k-a+o(1)\big)$$
as $k\to\infty$. 
For any real $b$ and $x>0$, let 
$$\ka:=\ka_b(x):=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big).$$
We have to show that 

for any real $b\in(-\infty,a)$, if $x>0$ is large enough and $f(k)\ge x$, then $k\ge\ka_b(x)$. 

Take any real $b\in(-\infty,a)$ and then take any $c\in(b,a)$. Let
$$f_c(k):=k\big(\ln k +\ln\ln k-c\big).$$ 
Let $x\to\infty$. Then 
$$\ln\ka=\ln x-\ln\ln x+o(1)\sim\ln x,$$
$$\ln\ln\ka=\ln\ln x+o(1),$$
$$\ln\ka+\ln\ln\ka=\ln x+o(1),$$
$$f_c(\ka_b(x))=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big)\big(\ln x-c+o(1)\big) \\
=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big)\Big(1-\frac{c-o(1)}{\ln x}\Big)\ln x \\ 
=x\Big(1+\frac{b-c-o(1)}{\ln x}\Big)<x$$ 
for large enough $x>0$. 
Now suppose that $f(k)\ge x$. Then $x\to\infty$ implies $k\to\infty$, because $f$ is bounded on any bounded subset of the set of all natural numbers. Therefore, for all large enough $x>0$ we have $f(k)\le f_c(k)$ and, because $f_c(k)$ is increasing in large enough $k$, we also have $f_c(k)\le f_c(\ka_b(x))$ if $k<\ka_b(x)$, so that 
$$f(k)\le f_c(k)\le f_c(\ka_b(x))<x,$$
which means that indeed, if $x>0$ is large enough and $f(k)\ge x$, then $k\ge\ka_b(x)$. 
