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Algebra - exercitiu rezolvat 1

Daca a, b, c > 0, atunci sa se arate ca:

$\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} \geq \frac{3}{2}$


Rezolvare

Notam

b + c = u
c + a = v
a + b = z

De aici obtinem:

$a = \frac{v + z - u}{2}$

$b = \frac{u + z - v}{2}$

$c = \frac{u + v - z}{2}$

Care inlocuite in inegalitate rezulta:

$\frac{v + z - u}{2u}+ \frac{u + z - v}{2v} + \frac{u + v - z}{2z} \geq \frac{3}{2}$

sau

$(\frac{v}{u} + \frac{u}{v}) + (\frac{z}{u} + \frac{u}{z}) + (\frac{v}{z} + \frac{z}{v}) \geq 6$

adevarata daca luam in considerare ca:

$\frac{a}{b} + \frac{b}{a} \geq 2, a, b > 0,$ cu egalitate pentru a = b


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