You could use np.einsum

y = a*b  np.einsum('i,i,j>j',a,b,b)
We can also perform a*b
and feed to einsum

y = a*b  np.einsum('i,j>j',a*b,b)
On the second approach, we can save some runtime by storing a*b
and reusing.
Runtime test 
In [253]: a = np.random.rand(4000)
In [254]: b = np.random.rand(4000)
In [255]: %timeit np.sum(np.outer(a*b, b), axis=0)
10 loops, best of 3: 105 ms per loop
In [256]: %timeit np.einsum('i,i,j>j',a,b,b)
10 loops, best of 3: 24.2 ms per loop
In [257]: %timeit np.einsum('i,j>j',a*b,b)
10 loops, best of 3: 21.9 ms per loop