Let $a_i>0$ for $i=1,...,n$. It is wellknown that $A\ge H$, where $A$ and $H$ are the arithmetic mean and harmonic mean of the vector $(a_i)$, respectively. Is any lower bound on $H/A$ known?

1$\begingroup$ there is no nonzero lower bound on $H/A$ that is independent of the $a_i$'s, the ratio can be arbitrarily close to zero. $\endgroup$ – Carlo Beenakker Jul 18 at 10:24

$\begingroup$ @CarloBeenakker Sure. I was thinking of something depending on some $a_i$'s (their maximum? their minimum?) and possibly $n$ and $A$. $\endgroup$ – Delio Mugnolo Jul 18 at 10:37

$\begingroup$ @CarloBeenakker Let me add that I'm aware of your answer here: mathoverflow.net/questions/195966/… $\endgroup$ – Delio Mugnolo Jul 18 at 11:03
If $(a_k)$ and $(b_k)$ are positive sequences of the same length, and $$0<m\le \frac{a_k}{b_k} \le M<\infty$$ $$A=\frac{m+M}{2},\ \ G=\sqrt{mM}$$ then $$(\Sigma{a_k}^2)(\Sigma{b_k}^2) \le (\frac{A}{G}\Sigma{a_kb_k})^2=\frac{A^2}{G^2}(\Sigma{a_kb_k})^2$$ This is a reverse of CauchySchwarz which follows from the trivial inequality $(M\frac{a_k}{b_k})(\frac{a_k}{b_k}m) \ge 0$ and the arithmeticgeometric mean inequality.
Applying this for $\sqrt{a_k}, \frac{1}{\sqrt{a_k}}$, we recover the inequality given in the answer by kodlu.

$\begingroup$ Sorry, I don't get it. Taking $a_k:=\sqrt{a_k}$ and $b_k:=\frac{1}{\sqrt{a_k}}$, wouldn't one find $(\sum a_k b_k)^2=n^2$? But this factor seems to be missing in Kodlu's formula. $\endgroup$ – Delio Mugnolo Jul 18 at 13:57

$\begingroup$ the arithmetic and harmonic means absorb each an $n$ $\endgroup$ – Conrad Jul 18 at 14:21

$\begingroup$ Sorry, stupid of me. Of course you're right, I answered too quickly. $\endgroup$ – Delio Mugnolo Jul 18 at 14:26

$\begingroup$ no problem  this way seems simpler and more direct than the Mitrinovic book (however that one and Mitrinovic's newer book about inequalities of functions, their integrals and their derivatives with Pecaric and Fink are utterly awesome and fundamental references) but that's probably also doable fairly elementarily $\endgroup$ – Conrad Jul 18 at 14:29
In Mitrinovic's Analytic Inequalities, published by Springer many years ago in the Grundlehren series, one finds, on page 79, the following inequality on the ratio $Q_{s,t}(a)=M_s(a)/M_t(a)$ of means of order $\infty<t<s<\infty:$ $$ Q_{s,t}(a)\leq \left( \frac{t(C^sC^t)}{ (st)(C^t1) } \right)^{1/s} \left( \frac{s(C^tC^s)}{ (ts)(C^s1) } \right)^{1/t},\quad st\neq 0, $$ where $C=\frac{\max_i a_i }{\min_i a_i}.$
One must be careful since the theorem is stated for weighted means, so I may have lost a factor of $n^a$ somewhere (I invite the OP to check) but if I have done the algebra correctly, this yields $$ \frac{A}{H}\leq \frac{(C+1)^2}{4C} $$