Is there a better way to find the minimum value of $\frac{1}{\sin \theta \cos \theta \left ( \sin \theta + \cos \theta \right )}$?

If $a^2+b^2=1$, then we can prove that $ab(a+b) \leq \frac{1}{\sqrt{2}}$ using: $ab \leq \frac{1}{2}(a^2+b^2)=\frac{1}{2}$ and $(a+b)^2=1+2ab \leq 2$. For equality to be attained, it is necessary that $a=b$.

Another approach:

Let $$\sin x+\cos x=p$$ then $p \in [-\sqrt{2} \: \sqrt{2}]$

Also $$\sin x \:\cos x=\frac{p^2-1}{2}$$ So we need to find Max and Min values of the function

$$g(p)=\frac{2}{p^3-p}=\frac{2}{f(p)}$$ in the interval $p \in [-\sqrt{2}\:\sqrt{2}]$

it will be evident from the Graph of $f(p)$ that in the above

interval $f(p)$ is Min at $p=-\sqrt{2}$ and is Max at $p=\sqrt{2}$

So

$\max(f(p))=\sqrt{2}$ and $\min(f(p))=-\sqrt{2}$

Thus

$\displaystyle \max(g(p))=\frac{2}{\sqrt{2}}=\sqrt{2}$ and Similarly $\displaystyle \min(g(p))=-\sqrt{2}$