inegalitate conditionata
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- Mesaje: 1493
- Membru din: Mar Oct 26, 2010 9:21 pm
- Localitate: ORADEA
inegalitate conditionata
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$.
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- Mesaje: 244
- Membru din: Sâm Oct 30, 2010 3:55 pm
- Localitate: Bradenton, Florida
Re: inegalitate conditionata
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.
|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.