<s>[INST] 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 $s = (a+b+c)/2$.
 [/INST] $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\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:

$$
\bbx{\sum_{\pars{a,b,c}\ \in\ T}
{2^{a} \over 3^{b}5^{c}} = {17 \over 21}}
$$
</s>