Does there exist a $C > 0$ such that $$ \sum_{n \geq 1} a_n \leq C \left( \sum_{n \geq 1} 2^n a_n^2 \right)^{1/4} \left( \sum_{n \geq 1} 2^{-n} a_n^2 \right)^{1/4} $$ for all $a_n \geq 0$ with finite rhs?
Progress. No single $\{a_n\}_n$ can be a counterexample, because $\sum_{n\geq1} 2^n a_n^2 < \infty$ implies $a_n^2 = O(2^{-n})$, from which the lhs is finite.
The sharp value of $C$ for sequences of the form $a_n = 2^{-\delta n}$ with $\delta > 1/2$ is $C = (71 + 17\sqrt{17})^{1/4} / (2\sqrt{2}) \approx 2.20457$. This is to say that $C > 1$, if it exists.
Works with $C=4$. Can be improved a bit if someone wants to.
Let $S=\sum_{n \geq 1} a_n$.
Case 1: There exists $N$ such that $a_N\ge S/4$. Then the right hand side of the above inequality is at least $$\Big(2^N (S/4)^2\Big)^{1/4}\Big(2^{-N} (S/4)^2\Big)^{1/4} = \frac{S}{4}$$
Case 2: There is no $N$ as above. Then let $N$ be the smallest integer such that $\sum_{n<N} a_n\ge S/4$. Observe that $\sum_{n< N} a_n\le S/2$ due to the minimality of $N$. Since also $a_N<S/4$, we have $\sum_{n>N} a_n\ge S/4$. By Cauchy-Schwarz,
$$\frac{S}{4}\le \sum_{n>N} 2^{-n/2} 2^{n/2}a_n \le \left( \sum_{n>N} 2^{-n}\right)^{1/2}\left( \sum_{n>N} 2^n a_n^2\right)^{1/2} $$ hence $$\left( \sum_{n } 2^n a_n^2\right)^{1/2} \ge 2^{N/2}\frac{S}{4} \tag1$$ Similarly, $$\frac{S}{4}\le \sum_{n<N} 2^{n/2} 2^{-n/2}a_n \le \left( \sum_{n<N} 2^{n}\right)^{1/2}\left( \sum_{n<N} 2^{-n} a_n^2\right)^{1/2} $$ hence $$\left( \sum_{n } 2^{-n} a_n^2\right)^{1/2} \ge 2^{-N/2}\frac{S}{4} \tag2$$ Multiply (1) and (2) and take square root.