Not sure how to approach this problem below. I tried assuming the conditional but that doesn't seem to get me anywhere.
If for a sequence of non-negative real numbers $\{a_n\}$, we have $$\sum{a_kb_k}$$ converges whenever $$\sum{b_k^2}$$ converges. then show that $$\sum{a_k^2}$$ converges
Suppose that the series $a_k^2$ is divergent, put $S_n=\sum_1^n a_k²$. Then $S_n \to +\infty$; hence we may suppose that $S_n>0$ for $n\geq 1$. Put $b_n=\frac{a_n}{S_n }$. Then as $$b_n^2=\frac{S_{n}-S_{n-1}}{S_n^2}\leq \int_{S_{n-1}}^{S_n}\frac{dt}{t^2}$$ we see that the series $b_n^2$ is convergent. The hypothesis gives now that the series $u_n=\frac{a_n^2}{S_n}=\frac{S_n-S_{n-1}}{S_n}$ is convergent. Hence $u_n\to 0$, and $\frac{S_{n-1}}{S_n}\to 1$. This imply that $u_n$ is equivalent to $v_n=\frac{S_n-S_{n-1}}{S_{n-1}}$, and hence (the two series are positive) the series $v_n$ is convergent. But $$v_n=\frac{S_n-S_{n-1}}{S_{n-1}}\geq \int_{S_{n-1}}^{S_n}\frac{dt}{t}$$ and this show that the series $v_n$ is divergent, a contradiction.
If $\sum a_kb_k$ and $\sum b_k^2$ converge, then their sum must also converge. Then using AM-GM: $$\sum b_k(a_k+b_k) \le \frac{1}{2}\sum b_k^2 +\frac{1}{2} \sum (a_k^2+2a_kb_k+b_k^2)<\infty \Rightarrow \sum a_k^2<\infty.$$
