Question

Daca x,y>0, atunci $\frac{ a^{2} }{x} + \frac{ b^{2} }{y} \geq \frac{(a+b)^{2}}{x+y}$ cu egalitate daca  $\frac{a}{x} = \frac{b}{y}$ . Deduceti inegalitatea
$\frac{ a^{2} }{x} + \frac{ b^{2} }{y} + \frac{ c^{2} }{z} \geq \frac{(a+b+c)^{2}}{x+y+z}$ , unde x,y,z>0 , a,b,c numere reale, cu egalitate daca
 $\frac{a}{x} = \frac{b}{y} = \frac{ c }{z}$

Answer 1

Aceste inegalitati provin din inegalitatea Cauchy-Bunyakovski-Schwarz. Le-am atasat.