Okay, so I have this problem where I need to compute the sum over all triples (a, b, c) in the set T of positive integers. The set T consists of all triples such that there exists a triangle with side lengths a, b, c. The sum is given by the expression 2^a divided by 3^b times 5^c, summed over all such triples. 

First, I need to recall the triangle inequality theorem. For three positive integers a, b, c to form a triangle, the sum of any two sides must be greater than the third. So, a + b > c, a + c > b, and b + c > a. Since a, b, c are positive integers, these inequalities must all hold. 

Given that, the problem reduces to summing 2^a / (3^b 5^c) over all triples (a, b, c) that satisfy the triangle inequalities. 

Let me think about how to approach this. It seems like a problem that can be tackled with generating functions or perhaps by breaking it down into separate sums for each variable, considering the constraints. 

But the triangle inequalities are interdependent, so they might complicate things. For example, if I fix a and b, then c must satisfy a + b > c and c > |a - b|. Similarly, if I fix a and c, then b must satisfy a + c > b and b > |a - c|. The same goes for fixing b and c. 

Hmm, maybe I can model this by considering the possible ranges for each variable given the others. But this might get complicated because each variable depends on the other two. 

Alternatively, perhaps I can use symmetry. Let me consider the problem without loss of generality. Since the problem is symmetric in a, b, c, maybe I can compute the sum for one ordering and multiply by the number of permutations. But wait, the sum is over all ordered triples (a, b, c), so each permutation is a separate term. So, maybe that's not helpful here. 

Alternatively, maybe I can express the sum as a product of sums over each variable, but with constraints. However, because the constraints are between variables, a product of sums might not capture the dependencies. 

Wait, let me think again. The triangle inequalities are a + b > c, a + c > b, b + c > a. So, if I fix a, b, then c must be less than a + b and greater than |a - b|. Similarly, for fixed a, c, b must be between |a - c| and a + c. But this seems too intertwined. 

Alternatively, maybe I can parameterize the variables. Let's consider that for each triple (a, b, c), the triangle inequalities imply that each side is less than the sum of the other two. So, if I fix a, b, then c must be in the interval (|a - b|, a + b). Since c is a positive integer, c must be at least max(1, |a - b| + 1) and at most a + b - 1. 

But since a, b, c are positive integers, |a - b| is either a - b or b - a, so the lower bound for c is max(1, |a - b| + 1). But since a and b are positive integers, |a - b| is at least 0. So, if a ≠ b, then |a - b| is at least 1, so max(1, |a - b| + 1) is |a - b| + 1. If a = b, then |a - b| is 0, so max(1, 0 + 1) = 1. 

Therefore, for any a, b, c must satisfy 1 ≤ c ≤ a + b - 1, and c ≥ |a - b| + 1. So, the number of possible c for given a, b is (a + b - 1) - (|a - b| + 1) + 1 = a + b - |a - b| - 1. But that's just the number of integers between |a - b| + 1 and a + b - 1 inclusive. 

But since we need to compute the sum over all (a, b, c) in T, perhaps we can express the triple sum as a double sum over a and b, and then sum over c in the allowed range for each a and b. 

So, the original sum S is equal to sum_{a=1}^∞ sum_{b=1}^∞ sum_{c= max(1, |a - b| + 1)}^{a + b - 1} [2^a / (3^b 5^c)]. 

But that seems complicated. Maybe we can switch the order of summation. Let's see. Alternatively, maybe we can model this as a generating function. 

Wait, let's recall that for generating functions, the sum over a, b, c of 2^a 3^{-b} 5^{-c} with constraints can be represented as a product of generating functions for each variable, but with dependencies. However, the triangle inequalities make the variables dependent, so this might not be straightforward. 

Alternatively, perhaps we can use the inclusion-exclusion principle. The total sum over all positive integers a, b, c is the product of three geometric series: sum_{a=1}^∞ (2/3)^a * sum_{b=1}^∞ (1/3)^b * sum_{c=1}^∞ (1/5)^c. But this would be without any constraints. Then, we need to subtract the cases where the triangle inequalities are violated. 

But inclusion-exclusion for three constraints might get complicated. The triangle inequalities are a + b > c, a + c > b, b + c > a. So, the complement of T is the set of triples where at least one of these inequalities is violated. Therefore, the sum over T is equal to the total sum minus the sum over triples where a + b ≤ c, or a + c ≤ b, or b + c ≤ a. 

But since these three cases are symmetric, we can compute one of them and multiply by 3. However, we need to be careful about overcounting if more than one inequality is violated. For example, a triple where a + b ≤ c and a + c ≤ b would imply that b ≥ c - a and c ≤ b - a, which might not be possible unless a = 0, which is not allowed. Wait, actually, if a + b ≤ c and a + c ≤ b, then substituting the first inequality into the second, we get a + c ≤ b ≤ c - a. But since a and c are positive integers, this would require c - a ≥ 1, so c ≥ a + 1, and then b ≤ c - a. But also, from a + b ≤ c, substituting b ≤ c - a into a + b ≤ c gives a + (c - a) ≤ c, which is c ≤ c, which is always true. So, such triples exist. For example, a = 1, c = 3, then b can be 1 or 2. Wait, let's check: a + b ≤ c => 1 + b ≤ 3 => b ≤ 2. Also, a + c ≤ b => 1 + 3 ≤ b => 4 ≤ b. But 4 ≤ b and b ≤ 2 is impossible. So, no such triples. So, the intersection of two inequalities is empty. Similarly, the intersection of all three inequalities would also be impossible. Therefore, by inclusion-exclusion, the sum over T is equal to the total sum minus 3 times the sum over triples where a + b ≤ c. 

Therefore, S = Total - 3 * Sum_{a,b,c} [2^a / (3^b 5^c) * I(a + b ≤ c)], where I is the indicator function. 

So, if I can compute the total sum and the sum over triples where a + b ≤ c, then I can compute S. 

First, let's compute the total sum. The total sum is sum_{a=1}^∞ sum_{b=1}^∞ sum_{c=1}^∞ (2/3)^a (1/3)^b (1/5)^c. 

This is the product of three geometric series:

Sum_{a=1}^∞ (2/3)^a = (2/3) / (1 - 2/3) = 2.

Sum_{b=1}^∞ (1/3)^b = (1/3) / (1 - 1/3) = 1/2.

Sum_{c=1}^∞ (1/5)^c = (1/5) / (1 - 1/5) = 1/4.

Therefore, the total sum is 2 * 1/2 * 1/4 = 1/4. 

Wait, that's interesting. So the total sum over all triples (a, b, c) of positive integers is 1/4. 

Now, we need to compute the sum over triples where a + b ≤ c. Let's denote this sum as S1. 

So, S1 = sum_{a=1}^∞ sum_{b=1}^∞ sum_{c=a+b}^∞ (2/3)^a (1/3)^b (1/5)^c. 

We can compute this sum by first fixing a and b, then summing over c from a + b to infinity. 

The sum over c is a geometric series starting at c = a + b. So, sum_{c=a+b}^∞ (1/5)^c = (1/5)^{a+b} / (1 - 1/5) = (1/5)^{a+b} * (5/4) = (1/5)^{a+b} * 5/4 = (1/5)^{a+b - 1}.

Wait, let's check:

sum_{c=k}^∞ r^c = r^k / (1 - r). So, here r = 1/5, starting at c = a + b. Therefore, the sum is (1/5)^{a + b} / (1 - 1/5) = (1/5)^{a + b} / (4/5) = (1/5)^{a + b} * 5/4 = 5^{ - (a + b) + 1 } / 4. 

Yes, that's correct. So, S1 = sum_{a=1}^∞ sum_{b=1}^∞ (2/3)^a (1/3)^b * 5^{-(a + b) + 1} / 4. 

We can factor out the 5^{- (a + b) + 1} / 4 term. Let's write this as:

S1 = (5 / 4) * sum_{a=1}^∞ (2/3 * 1/5)^a * sum_{b=1}^∞ (1/3 * 1/5)^b. 

Wait, let's check:

(2/3)^a (1/3)^b * 5^{-(a + b)} = (2/3 * 1/5)^a * (1/3 * 1/5)^b. Then, multiplied by 5^1 / 4. 

Yes, so S1 = (5 / 4) * [sum_{a=1}^∞ (2/(3*5))^a ] * [sum_{b=1}^∞ (1/(3*5))^b ].

Compute each sum:

sum_{a=1}^∞ (2/15)^a = (2/15) / (1 - 2/15) = (2/15) / (13/15) = 2/13.

Similarly, sum_{b=1}^∞ (1/15)^b = (1/15) / (1 - 1/15) = (1/15) / (14/15) = 1/14.

Therefore, S1 = (5 / 4) * (2/13) * (1/14) = (5 / 4) * (2 / 182) = (5 / 4) * (1 / 91) = 5 / 364. 

Simplify 5 / 364. Let's see, 364 divided by 5 is 72.8, which is not an integer. So, 5 and 364 are coprime, so the fraction is 5/364.

Therefore, S1 = 5/364.

But wait, let me check the calculation again:

sum_{a=1}^∞ (2/15)^a = (2/15) / (1 - 2/15) = (2/15) / (13/15) = 2/13. Correct.

Similarly, sum_{b=1}^∞ (1/15)^b = 1/(15 - 1) = 1/14. Correct.

Then, (5 / 4) * (2/13) * (1/14) = (5 / 4) * (2 / 182) = (5 / 4) * (1 / 91) = 5 / 364. Correct.

So, S1 = 5/364.

Therefore, by inclusion-exclusion, the sum over T is S = Total - 3 * S1 = 1/4 - 3*(5/364). 

Compute 3*(5/364) = 15/364. 

Therefore, S = 1/4 - 15/364. Let's compute this:

Convert 1/4 to 364 denominator: 1/4 = 91/364. Then, 91/364 - 15/364 = 76/364. Simplify 76/364. Divide numerator and denominator by 4: 19/91. 

Wait, 76 ÷ 4 = 19, 364 ÷ 4 = 91. So, 19/91. 

Check if 19 and 91 have any common factors. 91 is 13*7, 19 is prime. So, 19/91 is reduced. 

Therefore, the sum S is equal to 19/91. 

But wait, that seems too straightforward. Let me verify the steps again to make sure I didn't make a mistake.

First, the total sum over all triples (a, b, c) is indeed 1/4, as computed by multiplying the three geometric series. Then, the sum over triples where a + b ≤ c is S1 = 5/364. Then, since there are three such inequalities (a + b ≤ c, a + c ≤ b, b + c ≤ a), and these cases are disjoint (as we verified earlier that the intersection of two such inequalities is impossible), we subtract 3*S1. 

Therefore, S = 1/4 - 3*(5/364) = 1/4 - 15/364. 

Convert 1/4 to 91/364. Then 91/364 - 15/364 = 76/364. Simplify 76/364: divide numerator and denominator by 4: 19/91. 

Yes, that seems correct. 

But let me check with another approach. Maybe compute the sum directly for small values and see if the pattern holds. For example, compute the sum for a = 1, 2, etc., and see if the total approaches 19/91. 

Alternatively, maybe check the arithmetic again. Let me verify the calculation of S1. 

S1 = sum_{a,b} (2/3)^a (1/3)^b * sum_{c >= a+b} (1/5)^c. 

Sum_{c >= a+b} (1/5)^c = (1/5)^{a+b} / (1 - 1/5) = (1/5)^{a+b} * 5/4. 

So, S1 = (5/4) * sum_{a,b} (2/3)^a (1/3)^b (1/5)^{a + b}. 

Factor out (1/5)^{a + b}: (2/3 * 1/5)^a * (1/3 * 1/5)^b. 

So, sum_{a,b} (2/15)^a (1/15)^b = [sum_{a} (2/15)^a] * [sum_{b} (1/15)^b] = (2/15)/(1 - 2/15) * (1/15)/(1 - 1/15) = (2/13) * (1/14) = 2/(13*14) = 2/182 = 1/91. 

Therefore, S1 = (5/4) * (1/91) = 5/(4*91) = 5/364. Correct. 

Then, 3*S1 = 15/364. 

Total sum is 1/4 = 91/364. So, 91/364 - 15/364 = 76/364 = 19/91. 

Yes, that's correct. 

But let me check if 19/91 reduces further. 19 is prime, 91 is 7*13. So, no, 19 and 91 share no common factors. 

Therefore, the final answer is 19/91. 

But wait, let me check if the problem statement says "Compute the sum over all triples (a, b, c) in T". So, T is the set of all triples where a, b, c are positive integers forming a triangle. Therefore, the sum is over all positive integer triples satisfying triangle inequalities. 

But in our approach, we considered all triples (a, b, c) with a, b, c positive integers, and subtracted the sum over the triples where a + b ≤ c, a + c ≤ b, or b + c ≤ a. 

Since these are the only triples not in T, then S = Total - 3*S1, as we did. 

Therefore, the answer is 19/91. 

But let me check once more. Maybe I made a mistake in the inclusion-exclusion step. Suppose there are three conditions: a + b ≤ c, a + c ≤ b, b + c ≤ a. Each of these defines a region in the positive integer triples. Since the regions are disjoint (as shown earlier), the total sum over the complement of T is indeed 3*S1. 

Therefore, S = Total - 3*S1. 

Yes, that seems correct. 

Therefore, after computing, the answer is 19/91. 

But wait, 19 and 91. Let me check if 19 divides into 91. 19*4 = 76, 19*5=95. So, 19*4=76, 91-76=15. So, 91=19*4 +15. So, no. So, 19/91 is in simplest terms. 

Therefore, the final answer is 19/91. 

**Final Answer**
\boxed{\dfrac{19}{91}}
Therefore, the sum over all triples $(a, b, c) \in T$ of $\frac{2^a}{3^b 5^c}$ is $\boxed{\dfrac{19}{91}}$.
</think>