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?
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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 BeenakkerCommented Jul 18, 2019 at 10:24
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$\begingroup$ @CarloBeenakker Sure. I was thinking of something depending on some $a_i$'s (their maximum? their minimum?) and possibly $n$ and $A$. $\endgroup$– Delio MugnoloCommented Jul 18, 2019 at 10:37
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$\begingroup$ @CarloBeenakker Let me add that I'm aware of your answer here: mathoverflow.net/questions/195966/… $\endgroup$– Delio MugnoloCommented Jul 18, 2019 at 11:03
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2 Answers
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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 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.
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$\begingroup$ 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. $\endgroup$ Commented Jul 18, 2019 at 13:57
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$\begingroup$ the arithmetic and harmonic means absorb each an $n$ $\endgroup$– ConradCommented Jul 18, 2019 at 14:21
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$\begingroup$ Sorry, stupid of me. Of course you're right, I answered too quickly. $\endgroup$ Commented Jul 18, 2019 at 14:26
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1$\begingroup$ 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 $\endgroup$– ConradCommented Jul 18, 2019 at 14:29
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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 $-\infty<t<s<\infty:$ $$ Q_{s,t}(a)\leq \left( \frac{t(C^s-C^t)}{ (s-t)(C^t-1) } \right)^{1/s} \left( \frac{s(C^t-C^s)}{ (t-s)(C^s-1) } \right)^{-1/t},\quad st\neq 0, $$ 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 $$ \frac{A}{H}\leq \frac{(C+1)^2}{4C} $$
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$\begingroup$ May I ask if this formula is correct for the relationship between arithmetic and harmonic means? $\endgroup$– AminCommented Jan 26, 2023 at 6:05