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 $s = (a+b+c)/2$.
 A: 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.
$$\begin{align}
\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c} =
 &\left(\frac{2^2}{15^2} + \frac{2}{15}\right)\sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty \frac{2^{j+k}}{3^{i+k}5^{i+j}}\\
= & \frac{34}{225}\sum_{i=0}^\infty \left(\frac{1}{15}\right)^i \sum_{j=0}^\infty \left(\frac{2}{5}\right)^j \sum_{k=0}^\infty \left(\frac{2}{3}\right)^k\\
= &\frac{34}{225}\left(\frac{1}{1 - \frac{1}{15}}\right)
\left(\frac{1}{1 - \frac{2}{5}}\right)\left(\frac{1}{1 - \frac{2}{3}}\right)\\
= & \frac{17}{21}
\end{align}
$$
A: Naïve Solution
Note that $(a,b,c)\in\mathbb{Z}_{>0}^3$ is in $T$ iff $|b-c|<a<b+c$.  Therefore, if
$$S(p,q,r):=\sum_{(a,b,c)\in T}\,p^aq^br^c\,,$$
where $p,q,r\in\mathbb{C}$ with $|qr|<1$, $|rp|<1$, and $|pq|<1$, then
$$S(p,q,r)=\sum_{b=1}^\infty\,q^b\,\sum_{c=1}^\infty\,r^c\,\sum_{a=|b-c|+1}^{b+c-1}\,p^a\,.$$  Without loss of generality, we may assume that $p\neq \pm 1$.  That is,
$$S(p,q,r)=\sum_{b=1}^\infty\,q^b\,\sum_{c=1}^\infty\,r^c\,\left(\frac{p^{|b-c|+1}-p^{b+c}}{1-p}\right)\,.$$
Consequentlty,
$$S(p,q,r)=\small\frac{1}{1-p}\,\left(p\sum_{b=1}^\infty\,(qr)^b\,\sum_{c={b}}^\infty\,(pr)^{c-b}+p\,\sum_{b=1}^\infty\,(qr)^b\,\sum_{c=1}^{b-1}\,\left(\frac{p}{r}\right)^{b-c}-\sum_{b=1}^\infty\,(pq)^b\,\sum_{c=1}^\infty\,(pr)^c\right)\,.\tag{*}$$
If $p\neq r$, then
$$S(p,q,r)=\small\frac{1}{1-p}\,\Biggl(p\,\left(\frac{qr}{1-qr}\right)\,\left(\frac{1}{1-pr}\right)+p\,\sum_{b=1}^\infty\,(qr)^b\,\frac{\frac{p}{r}-\left(\frac{p}{r}\right)^{b}}{1-\frac{p}{r}}-\left(\frac{pq}{1-pq}\right)\,\left(\frac{pr}{1-pr}\right)\Biggr)\,.$$
Ergo, for $p\neq r$, we have
$$\begin{align}S(p,q,r)&=\frac{1}{1-p}\,\Biggl(p\,\left(\frac{qr}{1-qr}\right)\,\left(\frac{1}{1-pr}\right)+p\,\left(\frac{\frac{p}{r}}{1-\frac{p}{r}}\right)\,\left(\frac{qr}{1-qr}\right)\\&\phantom{aaaaa}-p\,\left(\frac{1}{1-\frac{p}{r}}\right)\,\left(\frac{pq}{1-pq}\right)-\left(\frac{pq}{1-pq}\right)\,\left(\frac{pr}{1-pr}\right)\Biggr)\,.\end{align}$$
Simplifying the expression, we get
$$S(p,q,r)=\frac{pqr\,(1+pqr)}{(1-qr)\,(1-rp)\,(1-pq)}\tag{#}$$
when $p\neq r$..
If $p=r$, then you can use continuity to conclude that (#) holds.  Alternatively, from (*), we have
$$S(p,q,p)=\small\frac{1}{1-p}\,\left(p\sum_{b=1}^\infty\,(pq)^b\,\sum_{c={b}}^\infty\,(p^2)^{c-b}+p\,\sum_{b=1}^\infty\,(b-1)\,(pq)^b-\sum_{b=1}^\infty\,(pq)^b\,\sum_{c=1}^\infty\,(p^2)^c\right)\,.$$
That is,
$$S(p,q,p)=\small\frac{1}{1-p}\,\Biggl(p\,\left(\frac{pq}{1-pq}\right)\,\left(\frac{1}{1-p^2}\right)+p\,\left(\frac{pq}{1-pq}\right)^2-\left(\frac{pq}{1-pq}\right)\,\left(\frac{p^2}{1-p^2}\right)\Biggr)\,.$$
Upon simplifying, we get
$$S(p,q,p)=\frac{p^2q\,(1+p^2q)}{(1-p^2)\,(1-pq)^2}\,,$$
which agrees with (#).
Now, in this particular problem, $p=2$, $q=\dfrac{1}{3}$, and $r=\dfrac{1}{5}$.  Therefore, by (#),
$$S\left(2,\frac13,\frac15\right)=\frac{17}{21}\,.$$  I was tempted to use Ravi's substitutions, but wanted to illustrate that a direct approach is not that bad.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{\pars{a,b,c}\ \in\ T}
{2^{a} \over 3^{b}5^{c}}:\ {\LARGE ?}\qquad
\\[2mm] T \equiv
\braces{\pars{a,b,c}\ \mid\ a,b,c \in \mathbb{N}_{\geq 1}\ \mbox{and}\ \verts{b - c} < a < b + c}}$

I'll evaluate
\begin{align}
&\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{\vphantom{\large A}\verts{b - c} < a < b + c}}
\\[5mm] = &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b < c}\bracks{\vphantom{\large A}c - b < a < c + b}
\\[2mm] + &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over \pars{yz}^{b}}
\bracks{b = c}\bracks{\vphantom{\large A}a < 2b}
\\[2mm] + &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c}
\\[5mm] = &\
\overbrace{\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}
\sum_{a = c - b + 1}^{c + b - 1}x^{a}}
^{\mbox{See expression}\ {\large\eqref{2}}}\ +\
\overbrace{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}}
\sum_{a = 1}^{2b - 1}x^{a}}
^{\mbox{See expression}\ {\large\eqref{3}}}
\\[2mm] + &\
\underbrace{\quad\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c}\quad}
_{\ds{\mbox{It's equal to the}\ first\ term\
\mbox{under the exchange}\ y \leftrightarrow z}}
\label{1}\tag{1}
\end{align}

The RHS first term becomes:
\begin{align}
&\bbox[#ffd,10px]{\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}
\sum_{a = c - b + 1}^{c + b - 1}x^{a}} =
\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}\,
x^{c - b + 1}\,{x^{2b - 1} - 1 \over x - 1}
\\[5mm] = &\
\sum_{b = 1}^{\infty}{1 \over y^{b}}
\,{x^{b} - x^{-b + 1} \over x - 1}
\sum_{c = b + 1}^{\infty}\pars{x \over z}^{c}
\\[5mm] = &\
{1  \over x - 1}\sum_{b = 1}^{\infty}
\,\bracks{\pars{x \over y}^{b} - {x \over \pars{xy}^{b}}}
{\pars{x/z}^{b + 1} \over 1 - x/z}
\\[5mm] = &\
{z  \over \pars{x - 1}\pars{z - x}}\sum_{b = 1}^{\infty}
\,\bracks{{x \over z}\pars{x^{2} \over yz}^{b} - {x^{2} \over z} \pars{1 \over yz}^{b}}
\\[5mm] = &\
{z  \over \pars{x - 1}\pars{z - x}}
\,\bracks{{x \over z}{x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} - {x^{2} \over z}{1/\pars{yz} \over 1 - 1/\pars{yz}}}
\\[5mm] = &\
\bbx{{x^{3} \over \pars{x - 1}\pars{z - x}\pars{yz - x^{2}}} -
{x^{2} \over \pars{x - 1}\pars{z - x}\pars{yz - 1}}}
\label{2}\tag{2}
\end{align}

The RHS second term in \eqref{1} is given by
\begin{align}
&\bbox[10px,#ffd]{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}}\,
\sum_{a = 1}^{2b - 1}x^{a}} =
\sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}\,
x\,{x^{2b - 1} - 1 \over x - 1}
\\[5mm] = &\
{1 \over x - 1}\bracks{%
\sum_{b = 1}^{\infty}\pars{x^{2} \over yz}^{b} -
x\sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}}
\\[5mm] = &\
{1 \over x - 1}\bracks{%
{x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} -
x\,{1/\pars{yz} \over 1 - 1/\pars{yz}}}
\\[5mm] = &\
\bbx{{1 \over x - 1}\pars{%
{x^{2} \over yz - x^{2}} -{x \over yz - 1}}}
\label{3}\tag{3}
\end{align}

The last term in \eqref{1} RHS is found, as pointed out above, with the exchange
$\ds{y \leftrightarrow z}$ in \eqref{2}.


The final result is given by

$$
\bbx{\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{\vphantom{\large A}\verts{b - c} < a < b + c}} =
{x\pars{x + yz} \over \pars{x - y}\pars{x - z}\pars{yz - 1}}}
$$

When $\ds{x = 2\,,\ y = 3}$ and $\ds{z = 5}$, the above expression is reduced to:

$$
\bbx{\sum_{\pars{a,b,c}\ \in\ T}
{2^{a} \over 3^{b}5^{c}} = {17 \over 21}}
$$
A: Not an answer but too much for a comment.
Case 1. $a=b=c, S_1=\sum_{n=1}^\infty{\left(\cfrac 2{15}\right)^n}$ That's the easy case;
Case 2. $\displaystyle a=b\ne c, S_2=\sum_{(a,c)\in U}\left(\cfrac{2^a}{3^a5^c}+\cfrac{2^a}{3^c5^a}+\cfrac{2^c}{3^a5^a}\right)=\sum_{i=1}^\infty\left(\cfrac{2^i}{3^i}\left(\sum_{j=1}^{2i-1, j\ne i}\cfrac 1{5^j}\right)+\cdots+\cdots\right)$, 
with $U$ the set of integer tuples that can form an isosceles triangle with the first entity the sides.
Case 3. $a< b < c$, $S_3=$sum of 6 terms (Permutation of 3) for each triple of $a,b,c $ satisfying $a+b>c$
$S_3$ needs further investigation. $S_1, S_2$ is almost there. Total=$S_1+S_2+S_3$ 
