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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?

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  • Your last paragraph is correct: Notice that if $a=b$ we have equality, so we can't make the LHS larger or the LHS smaller without the inequality being violated. – Jose27 Apr 27 '15 at 04:58
  • so there is no difference between sharp and tight, right? – chloe Jan 12 '23 at 22:57

2 Answers2

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Here is another definition of sharpness that demonstrates that the inequality is optimal (i.e., best-possible).

If $f,h:\mathbb{R}^n \longrightarrow \mathbb{R}$ and \begin{equation} f(x) \geq h(x),~\forall x \in S \subseteq\mathbb{R}^n, \tag{1} \label{ineq} \end{equation} then \eqref{ineq} is called sharp if $\exists s \in S$ such that $f(s) = h(s)$.

Why is this inequality the best possible inequality? Suppose that \eqref{ineq} is sharp as defined above. For contradiction, if $g:\mathbb{R}^n \longrightarrow \mathbb{R}$ is a function such that \begin{equation} f(x) > g(x) \ge h(x),~\forall x \in S \end{equation} then $f(s) > h(s)$, a contradiction. Thus, the inequality is the best-possible.

Pietro Paparella
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    So basically, if an inequality can be changed to a strict one, then it isn't sharp? – Alex Jones Apr 26 '17 at 20:31
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    @AlexanderJ93: Indeed; for instance, the inequality $|x| \geq -1$, which is valid for every real number $x$, is not sharp since $|x| > -1$ is also valid for all $x$. – Pietro Paparella Apr 26 '17 at 22:22
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    @AlexanderJ93: Notice also that $|x| > |x| - 1 \geq -1$ and the inequality $|x| - 1 \geq -1$ is (obviously) sharp. – Pietro Paparella Apr 26 '17 at 22:52
  • @PietroPaparella What about the specific case where $s$ is such that $f(s)=0$ and hence any multiplicative constant $C$ preserves sharpness? I have asked this follow-up question. – Morris Fletcher Sep 11 '23 at 19:02
  • @MorrisFletcher: nothing changes: suppose that $f(x) \ge g(x), \forall x \in S \subseteq \mathbb{R}^n$ and $\exists s \in S$ such that $f(s) = g(s) = 0$. If $\exists h: \mathbb{R}^n \longrightarrow \mathbb{R}$ such that $f(x) \ge h(x) > g(x), \forall x \in S$, then $0 = f(s) \ge h(s) > g(s) = 0$, a contradiction. – Pietro Paparella Sep 11 '23 at 21:25
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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.

Nate 8
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