How AM-GM is applied here

Question

I don't understand how AM-GM is applied in the last part of the picture. This is on the the $16^{th}$ page of the book in chapter $1$ about $AM-GM$.

Answer 1

We need to prove that: $$\frac{a+c}{b+c}+\frac{a+c}{a+d}\geq\frac{4(a+c)}{a+b+c+d}$$ or $$(a+b+c+d)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)\geq4,$$ which is true by AM-GM: $$(a+b+c+d)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)\geq2\sqrt{(b+c)(a+d)}\cdot\frac{2}{\sqrt{(b+c)(a+d)}}=4.$$ The second inequality we can prove by the same way.

I think a proof by C-S and after this by AM-GM is much more better.

Answer 2

$$\frac{\frac{a+c}{b+c}+\frac{a+c}{a+d}}2\geq\frac2{\frac{b+c}{a+c}+\frac{a+d}{a+c}}\quad(\because AM\ge HM)$$ Proceed similarly for the next term.