Also if $a_k, b_k >0, 0<m\le \frac{a_k}{b_k} \le M<\infty, k=1,..n$ and $A=\frac{m+M}{2}, G=\sqrt{mM}$, then we have a reverse to Cauchy-Schwarz given by:

$(\Sigma{a_k}^2)(\Sigma{b_k}^2) \le (\frac{A}{G}\Sigma{a_kb_k})^2=\frac{A^2}{G^2}(\Sigma{a_kb_k})^2$

which follows from the trivial inequality $(M-\frac{a_k}{b_k})(\frac{a_k}{b_k}-m) \ge 0, k=1,...n$ by the usual manipulations (sum and the AG means inequality).

Applying this for $\sqrt{a_k}, \frac{1}{\sqrt{a_k}}$ we recover indeed the inequality given in the earlier answer

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.