Q:

How to find minimum integer $k$ , $\frac{1}{e+1}+\ln(e+1)-1\leq \ln k$

I would appreciate if somebody could help me with the following problem:
Q: How to find minimum integer $k$ ?
$$\frac{1}{e+1}+\ln(e+1)-1\leq \ln k$$

A:

$$\frac{1}{e+1}+\ln(e+1)-1\leq \ln k$$
$$e^{\frac{1}{e+1}+\ln(e+1)-1}\leq k$$
$$k\geq \left(\frac{1}{e}+1\right) e^{\frac{1}{1+e}}$$
With a calculator you can evaluate right hand side as $\approx1.78998$ so smallest $k=2$.
To evaluate this without a calculator we can approximate $e$ as being between $2.5$ and $3$.
$$e^\frac{1}{1+e}<e^\frac{1}{1+2.5}<3^\frac{1}{1+2/5}=3^{\frac{2}{7}}$$
Let $a=3^\frac{2}{7}$ so $a^7=9$ and as $7^7=823543$ and $5^7=78125$ and $78125\times9<823543$ then $a<\frac75$ so $e^{\frac{1}{1+e}}<\frac75$.
So $$\left(\frac1e+1\right)e^\frac{1}{1+e}<\left(\frac25+1\right)\times\frac75=\frac{49}{25}<2$$
So the smallest $k\leq2$
We can show that $k>1$ as $\left(\frac1e+1\right)e^\frac{1}{1+e}>1$
Hence the smallest $k=2$.

