Theorem
For each sequence of positive real numbers one has the inequality
where denotes the natural base.
Proof We first consider the following qualitative problem.
AM-GM inequality directly implies
the left hand side diverges when n goes to infinity, which means we wouldn’t be able to bound directly with AM-GM. It makes sense as the bound get worse when we get being “highly unequal”, and do guarantees “highly unequal” . So the natural idea of modifying AM-GM is considering which being more equal.
\sum_{k=1}^{\infty}\left(a_{1} a_{2} \cdots a_{k}\right)^{1 / k} &=\sum_{k=1}^{\infty} \frac{\left(a_{1} c_{1} a_{2} c_{2} \cdots a_{k} c_{k}\right)^{1 / k}}{\left(c_{1} c_{2} \cdots c_{k}\right)^{1 / k}} \\ & \leq \sum_{k=1}^{\infty} \frac{a_{1} c_{1}+a_{2} c_{2}+\cdots+a_{k} c_{k}}{k\left(c_{1} c_{2} \cdots c_{k}\right)^{1 / k}} \\ &=\sum_{k=1}^{\infty} a_{k} c_{k} \sum_{j=k}^{\infty} \frac{1}{j\left(c_{1} c_{2} \cdots c_{j}\right)^{1 / j}}, \end{aligned}$$ by considering $$(c_1 c_2 \cdots c_j)^{1/j} = j + 1 \quad \text{for } j = 1, 2, \dots$$ $$s_k = c_k \sum_{j=k}^{\infty} \frac{1}{j(c_1 c_2 \cdots c_j)^{1/j}} = c_k \sum_{j=k}^{\infty} \frac{1}{j(j+1)} = \frac{c_k}{k},$$ we get $$\sum_{k=1}^{\infty}\left(a_{1} a_{2} \cdots a_{k}\right)^{1 / k} \leq \sum_{k=1}^{\infty}\left(1+\frac{1}{k}\right)^{k} a_{k},$$ together with $(1+\frac{1}{k})^{k}\leq e$, we proved the required inequality.