a+b+c=p+q+r=1

[math111]

Daca $a,b,c$ sunt numere reale ne-negative,iar $p,q,r$ parcurg $[0,1]$ si $a+b+c=p+q+r=1$,atunci $abc \le \dfrac{pa+qb+rc}{8}$.
( Bulgaria,1993)