<s>[INST] 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 :)
 [/INST] Using a result from Some Logarithmic Inequalities, E.R.Love :
$$
\forall x>0 \quad ; \quad \ln (1+x) > \frac{2x}{2+x} 
$$
Making the transformation $x \to \frac 1x$ gives us :
$$
\forall x > 0 \quad ; \quad \ln\left(1+\frac 1x\right) > \frac{\frac 2x}{2+\frac 1x} = \frac 2{2x+1}
$$
Thus if $g(x) = x^2 + \frac {2}{2x+1}$ then $g(x) \leq f(x)$ for $x>0$.
(Note : there may be easier bounds, but for this one the minima was the easiest to find)
The minimum of $g(x)$ is $\frac 54$ at $x = \frac 12$. We easily have $\frac {1+ 2 \ln 2}{4} \leq 1$ by using  $\ln 2 \leq 1$. Thus, we are done.
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