Prove that: $e-\ln(10)>\sqrt 2-1.$ The author's original inequality is as follows.

Prove that:
$$e-\ln(10)>\sqrt 2-1$$

Is there a good approximation for $$e-\ln 10?$$
Actually, I am also wondering that,
Where does $\sqrt 2-1$ come from?  Maybe, there exist relevant inequality?
My attempt:
$$-\ln 10>\sqrt 2-1-e\\
\ln 10<e+1-\sqrt 2\\
e^{e+1-\sqrt 2}>10$$
So, can we show that
$$e^x>10$$
when $x\ge e+1-\sqrt 2$?
I don't have a good idea.
 A: By Taylor's series at $196$:
$$\sqrt{2}=\frac{1}{10}\sqrt{200}<\frac{1}{10}(14+\frac{1}{7}-\frac{2}{14^3}+10^{-5})<1.4143\tag{1}$$
By Maclaurin's series:
$$e>1+1+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}>2,718\tag{2}$$
By using $\ln 10=-\ln(0.9)+2\ln3$:
$$-\ln(0.9)<\frac{1}{10}+\frac{1}{2\times10^2}+\frac{1}{3\times10^3}+10^{-4}<0.10544\tag{3}$$
and by using the Maclaurin series $\ln(\frac{1+x}{1-x})=2\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$, we have
$$2\ln 3<4(1+\frac{1}{12}+\frac{1}{80}+\frac{1}{448}+\frac{1}{2304}+10^{-4})<2.1972\tag{4}$$
Hence by $(3)$ and $(4)$,
$$\ln 10<0.10544+2.1972=2.30264\tag{5}$$
Finally by $(1)$, $(2)$ and $(5)$,
$$e+1-\sqrt{2}-\ln 10>2.718+1-1.4143-2.30264=0.00106>0.$$
A: I think this is numeric coincidence.
Anyway, let $a = e - 1 + \sqrt{2}$, expand $e^x$ to eighth terms:
$$ e^a > \sum_{n=0}^{8}\frac{a^n}{n!} > 10.$$ One still needs a calculator but logically this seems right.
A: First we show that $\log 10 < 76/33$, since
$$
\exp\Big(\frac{76}{33}\Big) > \sum_{n=0}^9 \frac{76^n}{33^n\,n!} = \frac{1316160031686037871}{131576558279536755}>10\,.
$$
Now it is readily seen that
$$
\frac{76}{33}<\frac{67997}{29520}\,,=1+1+\frac 1 2+\frac 1 6 + \frac 1{24}+\frac 1{120}+\frac 1{720}-\frac{17}{41}\,,
$$
hence
$$
\log 10<\frac{76}{33} < e - \frac{17}{41} = e+1-\frac{58}{41}\,,
$$
and finally (since $58/41 >\sqrt 2$)
$$
e-\log 10 > \sqrt 2 - 1\,.
$$
A: I hope the following will help.
$$\ln10=\ln2+\ln5=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+...\right)+$$
$$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+...\right)<$$$$<2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}+\frac{\left(\frac{1}{3}\right)^9}{7}+\frac{\left(\frac{1}{3}\right)^{11}}{7}+...\right)+$$
$$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}+\frac{\left(\frac{2}{3}\right)^{15}}{13}+...\right)=$$$$=2\left(\frac{1}{3}+\frac{\left(\frac{1}{3}\right)^3}{3}+\frac{\left(\frac{1}{3}\right)^5}{5}+\frac{\left(\frac{1}{3}\right)^7}{7}\cdot\frac{1}{1-\left(\frac{1}{3}\right)^2}\right)+$$
$$+2\left(\frac{2}{3}+\frac{\left(\frac{2}{3}\right)^3}{3}+\frac{\left(\frac{2}{3}\right)^5}{5}+\frac{\left(\frac{2}{3}\right)^7}{7}+\frac{\left(\frac{2}{3}\right)^9}{9}+\frac{\left(\frac{2}{3}\right)^{11}}{11}+\frac{\left(\frac{2}{3}\right)^{13}}{13}\cdot\frac{1}{1-\left(\frac{2}{3}\right)^2}\right)=$$
$$=\frac{8166549323}{3546482940}.$$
Can you end it now?
