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?
 A: 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
$$
\frac{A}{H}\leq \frac{(C+1)^2}{4C}
$$
A: 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.
