For a triangle with side lengths $a,$ $b,$ $c,$ let $s = \frac{a + b + c}{2},$ and let
\begin{align*}
x &= s - a = \frac{-a + b + c}{2}, \\
y &= s - b = \frac{a - b + c}{2}, \\
z &= s - c = \frac{a + b - c}{2}.
\end{align*}By the Triangle Inequality, $x,$ $y,$ and $z$ are all positive.  (This technique is often referred to as the Ravi Substitution.)  Note that
\begin{align*}
a &= y + z, \\
b &= x + z, \\
c &= x + y.
\end{align*}If $s$ is even, then $x,$ $y,$ and $z$ are all positive integers.  So, we can set $x = i,$ $y = j,$ and $z = k,$ which gives us the parameterization $(a,b,c) = (j + k, i + k, i + j).$

If $s$ is odd, then $x,$ $y,$ and $z$ are all of the form $n - \frac{1}{2},$ where $n$ is a positive integer.  So, we can set $x = i - \frac{1}{2},$ $y = j - \frac{1}{2},$ and $z = k - \frac{1}{2}.$  This gives us the parameterization $(a,b,c) = (j + k - 1, i + k - 1, i + j - 1).$

Thus, our sum is
\begin{align*}
\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c} &= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{2^{j + k - 1}}{3^{i + k - 1} 5^{i + j - 1}} \right) \\
&= \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \left( \frac{2^{j + k}}{3^{i + k} 5^{i + j}} + \frac{15}{2} \cdot \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \right) \\
&= \frac{17}{2} \sum_{i = 1}^\infty \sum_{j = 1}^\infty \sum_{k = 1}^\infty \frac{2^{j + k}}{3^{i + k} 5^{i + j}} \\
&= \frac{17}{2} \sum_{i = 1}^\infty \frac{1}{15^i} \sum_{j = 1}^\infty \left( \frac{2}{5} \right)^j \sum_{k = 1}^\infty \left( \frac{2}{3} \right)^k \\
&= \frac{17}{2} \cdot \frac{1/15}{1 - 1/15} \cdot \frac{2/5}{1 - 2/5} \cdot \frac{2/3}{1 - 2/3} \\
&= \boxed{\frac{17}{21}}.
\end{align*}
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a,$ $b,$ $c.$  Compute
\[\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}.\]
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\frac{17}{21}
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a,$ $b,$ $c.$  Compute
\[\sum_{(a,b,c) \in T} \frac{2^a}{3^b 5^c}.\]
test