Inegalitatea 3, exponentiala, x^y+y^x \geq 1
Moderators: Laurian Filip, Filip Chindea, Radu Titiu, maky, Cosmin Pohoata
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Inegalitatea 3, exponentiala, x^y+y^x \geq 1
Fie \( x, y>0 \). Aratati ca \( x^y+y^x>1 \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
-
Filip Chindea
- Newton
- Posts: 324
- Joined: Thu Sep 27, 2007 9:01 pm
- Location: Bucharest
Evident putem considera \( x, y < 1 \). Apoi substituim \( x = \frac{1}{1 + u} \), \( y = \frac{1}{1 + v} \), \( u, v > 0 \). Utilizand inegalitatea lui Bernoulli, deducem \( (1+u)^y < 1 + uy \) (deoarece \( y \in (0,1) \)). Deci \( x^y = \frac{1}{(1+u)^y} > \frac{1}{1 + uy} = \frac{1 + v}{1 + u + v} \), si analog \( y^x > \frac{1 + u}{1 + u + v} \). In concluzie, \( x^y + y^x > \frac{2 + u + v}{1 + u + v} = 1 + \frac{1}{1 + u + v} > 1 \).
Life is complex: it has real and imaginary components.