Question

Решите неравенство (картинка внизу)

Answer 1

$\frac{a^3+b^3+3b^2+3b+1}{a^2-ab-a+(b+1)^2}=\\\\\frac{a^3+(b+1)^3}{a^2-a(b+1)+(b+1)^2}=\\\\\frac{(a+(b+1))(a^2-a(b+1)+(b+1)^2}{a^2-a(b+1)+(b+1)^2}=a+(b+1)=a+b+1=\\\\4-\sqrt{3}+4+\sqrt{3}+1=9$