For positive real numbers (a), (b) satisfy (ln frac{{2 – 2ab}}{{a + b}} = 2ab + a + b – 2). Find the minimum value ({P_{min }}) of (P = a + 2b). – Math book

For positive real numbers $$a$$, $$b$$ such that $$\ln \frac{{2 – 2ab}}{{a + b}} = 2ab + a + b – 2$$. Find the smallest value $${P_{\min }}$$ of $$P = a + 2b$$.

A. $${P_{\min }} = \frac{{2\sqrt {10} – 3}}{2}$$. B. $${P_{\min }} = \frac{{3\sqrt {10} – 7}}{2}$$. C. $${P_{\min }} = \frac{{2\sqrt {10} – 1}}{2}$$. D. $${P_{\min }} = \frac{{2\sqrt {10} – 5}}{2}$$.
Condition: $$ab 0,\forall t > 0$$. Infer that the function $$f\left( t \right)$$ covariates on the interval $$\left( {0; + \infty } \right)$$.
Therefore, $$\left( * \right) \Leftrightarrow f\left[ {2\left( {1 – ab} \right)} \right] = f\left( {a + b} \right)$$$$\Leftrightarrow 2\left( {1 – ab} \right) = a + b$$$$\Leftrightarrow a\left( {2b + 1} \right) = 2 – b$$$$\Leftrightarrow a = \frac{{ – b + 2}}{{2b + 1}}$$.
We have $$P = a + 2b = \frac{{ – b + 2}}{{2b + 1}} + 2b = g\left( b \right)$$.
$$g’\left( b \right) = \frac{{ – 5}}{{{{\left( {2b + 1} \right)}^2}}} + 2 = 0$$$$\ Leftrightarrow {\left( {2b + 1} \right)^2} = \frac{5}{2}$$$$\Leftrightarrow 2b + 1 = \frac{{\sqrt {10} }}{2}\ )\( \Leftrightarrow b = \frac{{\sqrt {10} – 2}}{4}$$ .
From the variation table deduce $${P_{\min }} = g\left( {\frac{{\sqrt {10} – 2}}{4}} \right) = \frac{{2\sqrt { 10} – 3}}{2}$$.