Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x > 0, x \in \mathbb{R}: f(x) \geq \frac{1 + 2\ln(2)}{4}$ Given $f(x) = x^2 + \ln (1 + \frac{1}{x})$, prove that $\forall x\in(0,+\infty): f(x) \geq \frac{1 + 2\ln(2)}4$.
That would be same as proving that $x^2 + \ln (\frac{x + 1}{x}) - \frac{1 + \ln(4)}{4} \geq 0$, isn't?
I have also found the derivative $f'(x) = \frac{2x^3 + 2x^2 - 1}{x^2 + x}$ and that there is a solution of $f(x)$ in the interval of $\left[\frac12, \frac1{\sqrt 2}\right]$.
But from there I am stuck, I am not able to see how to link all the concepts.
Could someone give me a hint?
Thank you and Happy New Year 2021 :)

You are almost there.
Since $f(0) = f(+\infty) = +\infty$ and there is a unique solution $x_0 > 0$ such that $f'(x_0)=0$ (because $2x^2(1+x)$ is an increasing function of $x$) we only need to show that $f(x_0) > \frac{1+2 \ln 2}{4}$.
Since $x_0 \in \left(\frac 12, 1\right)$, then
$$f(x_0) = x_0^2 + \ln \left( 1+\frac{1}{x_0}\right)> \left(\frac 12\right)^2 + \ln \left( 1 + \frac{1}{1}   \right) = \frac{1+4\ln 2}{4}> \frac{1+2\ln2}{4}.\blacksquare$$
