= Solution
The <Chebyshev inequality> states that for a <random variable> $Z$ with finite <variance> and $a>0$,
$$
\mathbb P\bigl(|Z-\mathbb E[Z]|\geq a\bigr)\leq\frac{\operatorname{Var}(Z)}{a^2}.
$$
Write $\mu_n=\mathbb E[T]$. The ratio in the hypothesis presupposes $\mu_n>0$ for all sufficiently large $n$. On the event $T=0$, the deviation from the mean equals $\mu_n$. Thus the <Chebyshev inequality> with $a=\mu_n$ gives
$$
\boxed{\mathbb P(T=0)\leq\frac{\operatorname{Var}(T)}{\mathbb E[T]^2}\longrightarrow0.}
$$
This is the <second moment method>. It uses only the stipulated relative <variance> bound; no assumption that triangle indicators are independent is needed.
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