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O inegalitate draguta
Posted: Wed Nov 05, 2008 9:23 pm
by Claudiu Mindrila
Aratati ca \( 1<\frac{x}{x+z+t}+\frac{y}{x+y+t}+\frac{z}{x+y+z}+\frac{t}{x+y+t}<2, \forall x,y,z,t \in \mathbb{R}^*_+ \).
Marin Dolteanu, Concursul "Cristian S. Calude", 2001
Posted: Wed Nov 05, 2008 10:50 pm
by Marius Mainea
Pentru prima:
\( \frac{x}{x+z+t}>\frac{x}{x+y+z+t} \) si analoagele
Prin sumare se obtine apoi inegalitatea din stanga
Pentru a doua:
\( \frac{x}{x+z+t}<\frac{x+y}{x+y+z+t} \)
\( \frac{y}{x+y+t}<\frac{y+z}{x+y+z+t} \)
\( \frac{z}{x+y+z}<\frac{z+t}{x+y+z+t} \)
\( \frac{t}{y+z+t}<\frac{t+x}{x+y+z+t} \) si apoi prn adunare rezulta a doua inegalitate.