It is less or more the same as GH's proof, but let me explain how may one naturally come to such argument even without a priori knowing the constant. I do not refer here to other comments.
At first, by standard scaling argument, $a_i\in [1,A]$, $b_i\in [1,B]$. Let's try to estimate $\sum a_ib_i$ from below via $\sum a_i^p$ and $\sum b_i^q$. The easiest way is by summping up inequalities $a_ib_i\geq \alpha a_i^p+\beta b_i^q$ for some positive constants $\alpha$, $\beta$. This inequality may be rewritten as $1\geq \alpha x+\beta x^{-q/p}$, where $x=a_i^{p-1}/b_i$. Since the RHS is convex in $x$, it suffices to check for maximal and minimal possible values of $x$, which corresponds to minimal $a_i$ and maximal $b_i$ or viceversa. In other words, we need to check two inequalities $B\geq \alpha+\beta B^q$, $A\geq \alpha A^p+\beta$, which correspond to pairs $(a_i,b_i)=(1,B)$ and $(a_i,b_i)=(A,1)$. It is natural to take $\alpha$, $\beta$ so that both inequalities are equalities. This is $2\times 2$ system, we solve it to find $\alpha=(AB^q-B)/(B^qA^p-1)$, $\beta=(BA^p-A)/(B^qA^p-1)$. Now it remains to get $$ \sum a_ib_i\geq \alpha\sum a_i^p+\beta\sum b_i^q=\geq b_i^q\geq (\alpha p)^{1/p}(\beta q)^{1/q} (\sum a_i^p)^{1/p}(\sum b_i^q)^{1/q}.$$
