Answer by Aig for What is the manipulation used to transform $\sum_{cyc} \dfrac{a-bc}{a+bc}$ into $\sum_{cyc}\dfrac{bc}{a+bc}$?

It’s a standard technique while proving inequalities, to substitute a fraction with a complementary one in a sense that the sum of these fractions is $1$.

In this case $$\frac{a-bc}{a+bc}= \frac{a+bc-2bc}{a+bc}= \frac{a+bc}{a+bc}- \frac{2bc}{a+bc}=1- \frac{2bc}{a+bc}.$$

So the initial inequality turns into

$$1- \frac{2bc}{a+bc}+1- \frac{2ac}{b+ac}+1- \frac{2ab}{c+ab}\le\frac32$$

$$- \frac{2bc}{a+bc}- \frac{2ac}{b+ac}- \frac{2ab}{c+ab}\le-\frac32$$

$$ \frac{2bc}{a+bc}+\frac{2ac}{b+ac}+ \frac{2ab}{c+ab}\ge\frac32$$

$$ \frac{bc}{a+bc}+\frac{ac}{b+ac}+ \frac{ab}{c+ab}\ge\frac34.$$