1
$\begingroup$

I have an inequality with $a_n>0\forall n$ and $A_n\geq a_n\forall n$ that \begin{equation} \sum^N_{n=1}a_n\frac{a_n}{A_n}\geq \frac{(\sum^N_{n=1}a_n)^2}{\sum^N_{n=1}A_n} \end{equation} however I am interested in how tight this inequality is. Does anyone have an idea to analyse it?

So far my idea is equality holds when \begin{equation} \frac{a_1}{A_1}=\cdots=\frac{a_N}{A_N} \end{equation} but any idea to show how much gap when the strict $>$ holds? For example, I tried to calculate the difference between two sides of inequality, but it becomes very complicated.

$\endgroup$
  • $\begingroup$ It is sharp: Equality holds when $a_1 = \cdots = a_N = 0$. $\endgroup$ – Travis Oct 8 '14 at 10:55
  • $\begingroup$ You should include the assumptions about the range allowed for $a_n,A_N$. For example, it appears that $A_n$ should be positive. Perhaps there are additional "known" restrictions that affect how "tight" the inequality is. $\endgroup$ – hardmath Oct 8 '14 at 10:55
  • $\begingroup$ Yes you are right. I have edited the question. $\endgroup$ – user91232 Oct 8 '14 at 10:57
0
$\begingroup$

This is just the Cauchy–Schwarz inequality

https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality

It is well known (and follows from the proof) that equality occurs iff all numbers are proportional (this is the condition you wrote).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.