Credit for this solution goes to Ravi Boppana.
Lemma 1: If $r_1, r_2, \ldots , r_n$ are non-negative reals and $x_1, x_2, \ldots x_n$ are reals, then
$$\sum_{i, j}\min(r_{i}, r_{j}) x_{i}x_{j}\ge 0.$$
Proof: Without loss of generality assume that the sequence $\{r_i\}$ is increasing. For convenience, define $r_0=0$. The LHS of our inequality becomes
$\text{[ i r i x i 2 +2 i}$ b_i$. Similarly, we can assume that$a_j>b_j$. If$b_ib_j=0, then both sides are zero, so we may assume that b_i$and$b_j are positive. We then have from the definitions of r_i$and$x_i$ that
$\text{r i = a i b i -1 r j = a j b j -1 x i}$ = & $\text{b i x j}$ = & $\text{b j , .}$
This means that
$\text{(r i , r j ) x i x j = ( a i b i -1, a j b j -1 ) b i b j}$ & = $\text{(a i b j , a j b i )-b i b j}$ & = $\text{(a i b j , a j b i )- (a i a j , b i b j ) , .}$
This concludes the proof of Lemma 2. $\blacksquare$
We can then apply Lemma 2 and Lemma 1 in order to get that
$\text{i,j (a i b j , a j b i )- i, j (a i a j , b i b j )}$ & = $\text{i, j [ (a i b j , a j b i )- (a i a j , b i b j ) ]}$ & = $\text{i, j (r i , r j ) x i x j}$ & 0\, $\text{.}$
This implies the desired inequality.