Question

(x+y)(y+z)(z+x) $\geq$ 8xyz
(x+1)(y+1)(x+z)(y+z)$\geq$16xyz

Answer 1

la primul: x^2y+xyz+x^2z+xz^2+xy^2+zy^2+xyz+yz^2>=8xyz
x^2y+x^2z+xz^2+xy^2+zy^2+yz^2>=6xyz adevarat deoarece media aritmetica>=media geometrica
la al doilea: aceeasi idee. se desfac parantezele iar apoi media aritmetica>=media geometrica