How to weight loss easily?
Python 2, 51 bytes

lambda l:max((10**len(`l`)*a/b,s)for s,b,a in l)[1]

Does the obvious thing of finding the entry with the biggest ratio, but sidesteps the prohibition against floats by first multiplying the numerator by a huge input-dependent power of 10 before floor-dividing.
I'll prove this coefficient is big enough to make floor-division act the same difference as non-floor division.
Claim: If a1/b1 &gt; a2/b2, then floor(Na1/b1) &gt; floor(Na2/b2) for any N≥b1b2.
Proof: Note that a1/b1 - a2/b2 is a multiple of 1/b1b2, so a1/b1 - a2/b2 &gt; 0 implies that

a1/b1 - a2/b2
≥ 1/b1b2

Then, multiplying both sides by N,

Na1/b1 - Na2/b2
≥ N/b1b2 ≥ 1

So, since Na1/b1 and Na2/b2 differ by at least 1, their respective floors are distinct. ∎
Now, note that the product b1b2 has digit length at most their total digit length, which is less than the string length of the input. Since the input is in base 10, its suffices to use 10 to the power of its length N=10**len(`l`) to produce a number with more digits than it, guaranteeing the condition.