I am mainly motivated by the question regarding how to evaluate the goodness of an inequality $$f(x) \leq g(x), \text{ }\forall \text{ }x.$$
The first approach is using numerical evaluations to compare $f(x)$ and $g(x)$ by sampling many different $x$ values.
Another approach is to use the concept of 'sharp inequalities'. There are several questions already regarding sharp inequalities, e.g., this question and this one. It is clear that the inequality is sharp if $\exists$ $x_0$ such that $f(x_0) =g(x_0)$. Some people call such a bound `best possible'.
However, what if we could find another function $f_2$ such that $f(x) \leq f_2(x) \leq g(x), \text{ }\forall \text{ }x$, and $\exists$ $x_1 \not= x_0$ such that $f(x_1) < f_2(x_1)$? In this case, both $f$ and $f_2$ are sharp lower bounds for $g$; however, it seems that $f_2$ is 'better' than $f$, as $f(x_1) < f_2(x_1)$.
To define the notion of 'being better' above, we clearly cannot use the concept of 'sharpness', as both bounds are sharp. Then what concept should we use here? Should we say that $f_2$ is a 'tighter' lower bound than '$f$'? How is the concept of 'being tight' or 'tighter' formally defined?
Are there other ways to analyze whether $f$ is a good lower bound for $g$?
It would be much appreciated if references are also provided.
I don't know if there is a formal framework for analyzing inequalities this way. We usually want to prove sharp inequalities, but a sharp inequality is not always the "best possible" inequality. As a silly example, $$ |a\cdot b|\le 1\qquad(a,b\in\mathbb R^n, |a|,|b|\le 1) $$ is a sharp inequality because it is attained for various values of $a$ and $b$ in the description, but it's not the best possible. There is a better inequality which is of course $$ |a\cdot b|\le |a|\cdot|b|, $$ and this inequality is also sharp. It's better than the first inequality because $|a|\cdot|b|\le 1$.
If we have a non-negative quantity $f(a_1,a_2,\dots)$ which depends on the parameters $a_1,a_2,\dots$, we generally want to find the best expression $g(a_1,a_2,\dots)\ge 0$ which satisfies $$ f(a_1,a_2,\dots) \le g(a_1,a_2,\dots)\qquad(\text{for all $a_1,a_2,\dots$)} $$ in the sense that for any other quantity $g'(a_1,a_2,\dots)$ which also satisfies the above inequality for all the inputs $a_1,a_2,\dots$, we have $g(a_1,a_2,\dots)\le g'(a_1,a_2,\dots)$.
To connect with the earlier example, of course $g(a,b) = |a|\cdot|b|$ "beats" $g'(a,b) = 1$ for $|a|,|b|\le 1$, although both are sharp upper bounds for $f(a,b) = |a\cdot b|$.