Square of a real number is always non-negative. So,

$(a+b-c)^2 \geq 0$

$\implies a^2+b^2+c^2 +2ab-2bc-2ac \geq 0$

$\implies a^2+b^2+c^2 \geq 2bc + 2ac - 2ab$

$\implies \frac{1}{abc}(a^2+b^2+c^2) \geq \frac{1}{abc}(2bc + 2ac - 2ab)$

$\implies \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}\geq \frac{2}{a}+\frac{2}{b}-\frac{2}{c}$