By squaring, we obtain:

$$
\frac{a b}{c}+\frac{b c}{a}+\frac{a c}{b}+2(a+b+c) \geqslant K^{2}(a+b+c)
$$

According to the turnstile lemma with $x=\sqrt{\frac{a b}{c}}, y=\sqrt{\frac{b c}{a}}$ and $z=\sqrt{\frac{a c}{b}}$, we have $\frac{a b}{c}+\frac{b c}{a}+\frac{a c}{b} \geqslant a+b+c$.

Thus, $3 \geqslant K^{2}$. By taking for example $a=b=c=1, K=\sqrt{3}$ is not only an upper bound constant but also achievable. This is therefore the answer to this exercise.
Find the largest constant $K$ such that for all positive real numbers $a, b$, and $c$, we have

$$
\sqrt{\frac{a b}{c}}+\sqrt{\frac{b c}{a}}+\sqrt{\frac{a c}{b}} \geqslant K \sqrt{a+b+c}
$$