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