Show $ f(m) = \left( 1 - \frac{1}{m^2} \right)^{m(m-1)/2} $ is lower-bounded by $\frac{1}{2}$ for $m \ge 1$ It is also possible to assume $m$ is an integer if this is useful. 
One way I can see to solve the problem is to show 
(1) $f(m)$ is monotonically decreasing, which can be done either by showing $f'(m) \leq 0$ or showing $f(m+1) \leq f(m)$ for all $m \ge 1$ .
(2) The limit as $m \to \infty$ of $f(m)$ is $\frac{1}{\sqrt{e}} > \frac{1}{2}$.
Showing $(2)$ is simple using $\lim_{n\to \infty} \left(1 + \frac{1}{n} \right)^n = e$. Showing $(1)$ is quite annoying using the derivative, I've reduced showing $(1)$ using the other approach to showing $\left(1 - \frac{1}{(m+1)^2} \right)^{m+1} \leq \left(1 - \frac{1}{m^2} \right)^{m-1}$.

$$
\begin{align}
\frac{\left(1-\frac1{m^2}\right)^{m-1}}{\left(1-\frac1{(m+1)^2}\right)^{m+1}}
&=\left(\frac{m^2}{m^2-1}\frac{(m+1)^2}{(m+1)^2-1}\right)\left(\frac{m^2-1}{m^2}\frac{(m+1)^2}{(m+1)^2-1}\right)^m\tag1\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(\frac{(m-1)(m+1)^3}{m^3(m+2)}\right)^m\tag2\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(1-\frac{2m+1}{m^3(m+2)}\right)^m\tag3\\
&\ge\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(1-\frac{2m+1}{m^2(m+2)}\right)\tag4\\
&=\left(\frac{m(m+1)}{(m+2)(m-1)}\right)\left(\frac{(m-1)(m^2+3m+1)}{m^2(m+2)}\right)\tag5\\
&=\frac{(m+1)(m^2+3m+1)}{m(m+2)^2}\tag6\\
&=\frac{m^3+4m^2+4m+1}{m^3+4m^2+4m}\tag7\\[9pt]
&\gt1\tag8
\end{align}
$$
Explanation:
$(1)$: move a factor from the numerator and denominator to the front
$(2)$: factor
$(3)$: write second factor as $1-x$
$(4)$: Bernoulli's Inequality
$(5)$: factor
$(6)$: combine factors
$(7)$: multiply
$(8)$: numerator is $1$ greater than the denominator
Thus, the sequence is decreasing.
