[tex]a,b,c \geq 0 \Rightarrow a+b+c \geq 3 \sqrt[3]{abc} \Leftrightarrow \Big(\frac{n}{3} \Big)^3 \geq abc.~Deci~maximul~ \\ \\ produsului~abc~este~ \frac{n^3}{27}~si~se~realizeaza~cand~a=b=c~(inegalitatea \\ \\ mediilor). \\ \\ De~altfel...fara~a~utiliza~inegalitati~celebre~avem~identitatea: \\ \\ x^3+y^3+z^3 -3xyz = \frac{1}{2} \Big(x+y+z \Big) \Big( (x-y)^2+(y-z)^2+(z-x)^2\Big). [/tex]
[tex]Alegand~x,y,z \geq 0~avem~x+y+z \geq 0.~Si~cum~(x-y)^2+(y-z)^2+ \\ \\ +(z-x)^2 \geq 0 ~rezulta~x^3+y^3+z^3-3xyz \geq 0 ,~cu~egalitate \Leftrightarrow \\ \\ x+y+z=0~(care~implica~x=y=z=0)~sau~x=y,~y=z,~z=x, \\ \\ (adica~x=y=z). \\ \\ Cazul~de~fata~este~x= \sqrt[3]{a},~y= \sqrt[3]{b}~si~z= \sqrt[3]{c}. [/tex]
