Lower bound of Prime-counting function. Assume that $t \geq n \geq 4$ and $\tau=\lceil tn\ln(tn) \rceil$. Then I know that the for the number of primes $\leq \tau$, $\pi(\tau)$ it holds that
$$
\pi(\tau) \geq \dfrac{\tau}{\ln \tau}
$$
I then need to show that
$$
\dfrac{\tau}{\ln \tau} \geq \dfrac{tn e}{e+1}
$$
How do I do this?
I know that
\begin{align*}
\tau &= \lceil tn\ln(tn) \rceil \geq tn\ln(tn) \geq tn \ln(4^2) \geq tn e\\
\ln\tau &= \ln\lceil tn\ln(tn) \rceil \geq \ln\lceil 4^2\ln(4^2) \rceil = 4 \geq e+1
\end{align*}
However, I am not sure how I can use this.
 A: Consider the function $f(x)=\frac {x}{\ln x}$. It can be easily shown that, $f'(x)>0$ for $x>e$. Now let $x= tn$. Then we have, since $\tau>x\ln x$ and $\tau, x\ln x\geq 16>e$, $f(\tau)>f(x\ln x)$. Now, proceeding through this, it suffices to show:
$$\frac {x\ln x}{\ln(x\ln x)}\geq \frac {xe}{e+1}$$
Canceling $x$ and cross multiplying (can be done as $x\geq 16$):
$$\ln x (1+\frac {1}{e})\geq \ln(x\ln x)$$
Taking antilog, this reduces to:
$$x^{\frac 1e}\geq \ln x$$
Now verify that for $x=16$ inequality holds, and since $x^{\frac 1e}-\ln x$ is an increasing function for $x>e^e$, inequality is true for subsequent values as well.
Since this last result is true, it follows that the initial inequality is true as well, as each step can be retraced to the last.
A: Let $u=tn$, of which we know $u\ge 16$.
Then $\tau\ge u\ln u$, but also $\tau< u\ln u+1=c u\ln u$, where $c$ is a number only slightly greater than $1$. Then $\ln(\tau)< \ln u+\ln\ln u+\ln c$,
so
$$\frac\tau{\ln \tau}>\frac{u\ln u}{\ln u+\ln\ln u+\ln c}.$$
Hence it suffices to show that
$$\frac{\ln u}{\ln u+\ln\ln u+\ln c}\stackrel?>\frac e{e+1} $$
or equivalently, after taking reciprocals and subtracting $1$
$$ \frac{\ln\ln u+\ln c}{\ln u}\stackrel?<\frac {1}e$$
So you probably want to find a bound for $\frac {\ln x}x$ when $x=\ln u \ge\ln 16$, and also a verify that $c$ is really not much greater than $1$, i.e., $\ln c$ really not much graeter than $0$. Can you take it from here?
