"Sharp" Inequalities When we say that an inequality is sharp, does it mean that it is "the best" inequality we can get between the two quantities involved?
For example, I read that we would say that the inequality
$$
\frac{a^2+b^2}{2}\geq ab
$$
is sharp, but wouldn't $\frac{a^2+b^2}{2}$ on the RHS be sharper than $ab$?
Do we just mean that we can't multiply the RHS of $\cdot\geq\cdot$ by a constant $>1$ (or equivalently that we can't multiply the LHS by a constant in $[0,1)$)? So that would be a "best" inequality in this sense?
 A: As was said, many times bounds are not as strong as they theoretically could be. In math, this comes because a specific quality is hard to compute. A basic example is computing world population-- you could say "The population of the world is greater than 3 billion." but this inequality might not be very good.
So by saying an inequality is sharp, it means that a theoretical maximum has been reached and there's no room for improvement for that particular inequality. However, there could of course be other inequalities with the same LHS or RHS.
A: Here is another definition of sharpness that demonstrates that the inequality is optimal (i.e., best-possible).
If $f,g:\mathbb{R}^n \longrightarrow \mathbb{R}$ and
\begin{equation} 
f(x) \geq g(x),~\forall x \in S \subseteq\mathbb{R}^n, \tag{1} \label{ineq}
\end{equation}
then \eqref{ineq} is called sharp if there is an element $\hat{x} \in S$ such that $f(\hat{x}) = g(\hat{x})$.
Why is this inequality the best possible inequality? Suppose that \eqref{ineq} is sharp as defined above. For contradiction, if $h:\mathbb{R}^n \longrightarrow \mathbb{R}$ is a function such that
\begin{equation} 
f(x) \ge h(x) > g(x),~\forall x \in S  
\end{equation}
then, in particular, $f(\hat{x}) > g(\hat{x})$, a contradiction. Thus, the inequality is the best-possible.
