We have $F(x) = x + \ln \left(\frac{1+2e^{-2x}}{1+e^{-x}}\right)$ represents curve $C$ and a line $d: y=x$
In an exercise it is required to study the relative position of these two functions, I know we have to subtract them from each other but I can't continue it
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$\begingroup$Let us ask for what $x$ the curve $y=F(x)$ is above the line $y=x$.
This happens precisely when $$x +\ln\left( \frac{1+2e^{-2x}}{1+e^{-x}} \right)\gt x.\tag{A}$$
The inequality (A) holds precisely if $$\ln\left( \frac{1+2e^{-2x}}{1+e^{-x}} \right)\gt 0.\tag{B}$$ The inequality (B) holds precisely if $$ \frac{1+2e^{-2x}}{1+e^{-x}}\gt 1.\tag{C}$$ The inequality (C) holds precisely if $$1+2e^{-2x} \gt 1+e^{-x}.\tag{D}$$ (Here we used the fact that $1+e^{-x}$ is always positive, so multiplying both sides of (C) by $1+e^{-x}$ yields an equivalent inequality.)
The inequality (D) is equivalent to $2e^{-2x}\gt e^{-x}$, which in turn is equivalent to the inequality $2\gt e^x$ (we multiplied both sides by $e^{2x}$).
Finally, the inequality $e^x \lt 2$ holds precisely when $x\lt \ln 2$.
So at any $x\lt \ln 2$, the curve $y=F(x)$ lies above the line $y=x$. When $x=\ln 2$, the curve and the line meet. And when $x\gt \ln 2$, the curve $y=F(x)$ is below the line $y=x$.
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