<s>[INST] Prove that $\frac{7}{12}<\ln 2<\frac{5}{6}$ using real analysis I studying in Real Analysis 2, but I have no idea how to solve this problem. My guess is to use Mean Value Theorem or a similar theorem? Could any one help me?
Thanks.
 [/INST] Consider the following inequalities from Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$
as suggested by @labbhattacharjee
$$\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx>0$$
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx>0$$
The integrals are positive because the integrands are positive for $0<x<1$.
Let us evaluate the first integral
$$\begin{align}
\frac{1}{2}\int_{0}^{1}\frac{x^2(1-x)}{1+x}dx
&=
\frac{1}{2}\int_{0}^{1}\frac{x^2-x^3}{1+x}dx \\
&=
\frac{1}{2}\int_{0}^{1}\left(-x^2+2x-2+\frac{2}{1+x}\right)dx\\
&=
\left.\frac{1}{2}\left(-\frac{x^3}{3}+x^2-2x+2\log(1+x)\right) \right|_{0}^{1}\\
&=
\frac{1}{2}\left(-\frac{1}{3}+1-2+2\log(2)\right) \\
&=
\log(2)-\frac{2}{3}>0 \\
\end{align}
$$
Similarly,
$$\frac{1}{2}\int_{0}^{1}\frac{x^5(1-x)}{1+x}dx=\frac{7}{10}-\log(2)>0$$
Combining both results,
$$\frac{2}{3}-\log(2)<0<\frac{7}{10}-\log(2)$$
so
$$\frac{2}{3}<\log(2)<\frac{7}{10}$$
To solve your problem we shall compare $\frac{2}{3}$ against $\frac{7}{12}$ and $\frac{7}{10}$ against $\frac{5}{6}$
On the one hand,
$$21<24$$
$$7·3<2·12$$
$$\frac{7}{12}<\frac{2}{3}$$
On the other hand,
$$42<50$$
$$7·6<5·10$$
$$\frac{7}{10}<\frac{5}{6}$$
Finally,
$$\frac{7}{12}<\frac{2}{3}<\log(2)<\frac{7}{10}<\frac{5}{6}$$
so
$$\frac{7}{12}<\log(2)<\frac{5}{6}$$
</s>