Question

Problema 181.Dau 50 puncte!!

Answer 1

Notam cu x=ac²b;y=ba²c;z=cb²a
Folosim o inegalitate uzuala si anume
logx y+㏒y x≥2;㏒y z+㏒z y≥2;㏒z x+logₓ z≥2
Insumind cele 3 relatii obtinem:
logₓy+logₓz+logy x+logy z+logz y+logz x=logₓyz+logy xz+logz xy≥6
logₓ(yz)=logac²b(ba²ccb²a)=1+2logac²b(ab);
in mod analog avem:
logy(xz)=1+2logba²c(bc);
logz(xy)=1+2logcb²a(ca);
∑=3+2(logac²b ab +logba²c bc+logcb²a ca)≥6
2(...........................)≥3 de unde
logac²b(ab)+logba²c(bc)+logcb²a(ca)≥3/2

Banuiesc ca problema e din gazeta matematica fiindca are un grad ridicat de dificultate,deci in mod normal punctele puse ar fi  trebuit sa fie mai multe!!!
Bafta!!!