TITLE: On inequality involving logarithm
QUESTION [0 upvotes]: I have been researching in order to do a paperwork about inequalities for my university and I ran into this inequality from a book I have been trying to understand without results.
It says:

For every $t\geq 1$ we have that
\begin{equation}
\ln(t)\leq \frac{(t+1)(t^3-1)}{3t(t^2+1)}
\end{equation}

The book also sais that to prove it, it suffices to show that the right side has a larger derivative for $t>1$. But I dont see how and why would this imply the inequality, also I have been trying to prove that the derivative is larger and I also cant..
Can someone help?
The book: https://www.jstor.org/stable/2321420

REPLY [0 votes]: The right-hand side $f(t)$ of the inequality has a zero at $t=1$, thus $\ln(1) = f(1)$.  If you manage to show that
$$f'(t) \geqslant \ln't =\frac1t \quad\text{ for all } t\geqslant 1 $$
then it follows that $f(t)\geqslant \ln(t)$ for all $t\geqslant 1$ because $f$ grows faster than $\ln$. (And the functions that are involved are smooth enough).
And similarly for $t\leqslant 1$.
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