The given equation rewrites as $$a_{i+2}\sqrt{2}-a_i=a_{i+1}\sqrt{3}\iff 2a_{i+2}^2+a_i^2-2\sqrt{2}a_ia_{i+2}=3a_{i+1}^2.$$From here it follows that $$2\sqrt{2}a_ia_{i+2}=2a_{i+2}^2+a_i^2-3a_{i+1}^2 \iff 2\sqrt{2}\sum a_ia_{i+2}=2\sum a_{i+2}^2+\sum a_i^2-3\sum a_{i+1}^2.$$ However, $RHS = 2\sum a_i^2+\sum a_i^2-3\sum a_i^2=0.$ Hence, $2\sum a_ia_{i+2}=0$ and we are done.