Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle 
Let $T$ be the set of all triples $(a,b,c)$ of positive integers for which there exist triangles with side lengths $a$, $b$, $c$. Express 
  $$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c}$$
  as a rational number in lowest terms

I really don't know to start with this question. My solution: I have written equation of solution of triangle in form of $a$, $b$, and $c$; i.e, $s/(s-a)$, where $s = (a+b+c)/2$.

The main difficulties of this problem is how to handle the triangle inequalities:
$$a+b > c, b+c > a, c+a > b$$
The beauty of Ravi substitution (mentioned by @maveric in comment) 
$$a = v + w, b = u + w, c = u + v$$
is under Ravi substitution,

$a, b, c$ satisfies the triangle inequalities if and only if $u,v,w$ are positive numbers.

If one want to run through all integer combinations of $a,b,c$ which give a triangle,
one just run $u,v,w$ through all positive integer combinations 
or positive half-integer combinations 
$$(u,v,w) = \text{ one of }\begin{cases}
(i+1, j+1, k+1 )\\
\\
(i+\frac12+i,j+\frac12,k+\frac12)
\end{cases}
\quad\text{ for }i,j, k\in \mathbb{N}$$
This transform the horrible sum to one over geometric series.
$$\begin{align}
\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c} =
 &\left(\frac{2^2}{15^2} + \frac{2}{15}\right)\sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty \frac{2^{j+k}}{3^{i+k}5^{i+j}}\\
= & \frac{34}{225}\sum_{i=0}^\infty \left(\frac{1}{15}\right)^i \sum_{j=0}^\infty \left(\frac{2}{5}\right)^j \sum_{k=0}^\infty \left(\frac{2}{3}\right)^k\\
= &\frac{34}{225}\left(\frac{1}{1 - \frac{1}{15}}\right)
\left(\frac{1}{1 - \frac{2}{5}}\right)\left(\frac{1}{1 - \frac{2}{3}}\right)\\
= & \frac{17}{21}
\end{align}
$$
