# Bound on the ratio of harmonic and arithmetic mean

Let $$a_i>0$$ for $$i=1,...,n$$. It is well-known 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?

• 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. – Carlo Beenakker Jul 18 '19 at 10:24
• @CarloBeenakker Sure. I was thinking of something depending on some $a_i$'s (their maximum? their minimum?) and possibly $n$ and $A$. – Delio Mugnolo Jul 18 '19 at 10:37
• @CarloBeenakker Let me add that I'm aware of your answer here: mathoverflow.net/questions/195966/… – Delio Mugnolo Jul 18 '19 at 11:03

If $$(a_k)$$ and $$(b_k)$$ are positive sequences of the same length, and $$0 $$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$$ This is a reverse of Cauchy-Schwarz which follows from the trivial inequality $$(M-\frac{a_k}{b_k})(\frac{a_k}{b_k}-m) \ge 0$$ and the arithmetic-geometric mean inequality.
Applying this for $$\sqrt{a_k}, \frac{1}{\sqrt{a_k}}$$, we recover the inequality given in the answer by kodlu.
• Sorry, I don't get it. Taking $a_k:=\sqrt{a_k}$ and $b_k:=\frac{1}{\sqrt{a_k}}$, wouldn't one find $(\sum a_k b_k)^2=n^2$? But this factor seems to be missing in Kodlu's formula. – Delio Mugnolo Jul 18 '19 at 13:57
• the arithmetic and harmonic means absorb each an $n$ – Conrad Jul 18 '19 at 14:21
In Mitrinovic's Analytic Inequalities, published by Springer many years ago in the Grundlehren series, one finds, on page 79, the following inequality on the ratio $$Q_{s,t}(a)=M_s(a)/M_t(a)$$ of means of order $$-\infty $$Q_{s,t}(a)\leq \left( \frac{t(C^s-C^t)}{ (s-t)(C^t-1) } \right)^{1/s} \left( \frac{s(C^t-C^s)}{ (t-s)(C^s-1) } \right)^{-1/t},\quad st\neq 0,$$ 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 $$\frac{A}{H}\leq \frac{(C+1)^2}{4C}$$