inegalitate conditionata

[mihai miculita]

Aratati ca, oricare ar fi numerele reale $x$ si $y$, care indeplinesc conditia: $x^2+y^2=2$, are loc inegalitatea: $-2$$\le$$xy(x+y)$$\le 2$.

[Virgil Nicula]

2|xy| <sau= x^2+y^2=2 => |xy| <sau= 1

|x+y|^2 <sau= le 2(x^2+y^2)=4 => |x+y| <sau= 2

Produsul celor doua inegalitati subliniate conduce la |xy(x+y| <sau= 2 , adica -2 <sau= xy(x+y) <sau= 2.