1. Zadatak

Vrijedi

$$y=\frac{(a+c-b)(a+b-c)}{(a+b+c)(b+c-a)} =\frac{(a-(b-c))(a+(b-c))}{((b+c)+a)((b+c)-a)} =\frac{a^2-(b-c)^2}{(b+c)^2-a^2}$$

pa je

$$y+1=\frac{a^2-(b-c)^2+(b+c)^2-a^2}{(b+c)^2-a^2} =\frac{a^2-b^2+2bc-c^2+b^2+2bc+c^2-a^2}{(b+c)^2-a^2} =\frac{4bc}{(b+c)^2-a^2}.$$

Vrijedi

$$x+1=\frac{b^2+c^2-a^2+2bc}{2bc} =\frac{(b+c)^2-a^2}{2bc}$$

te je

$$(x+1)(y+1) =\frac{(b+c)^2-a^2}{2bc}\cdot\frac{4bc}{(b+c)^2-a^2} =2.$$