begood Posted January 5, 2013 Report Share Posted January 5, 2013 (edited) Se da :x,y,z > 0x + y + z = xy + yz + zxSa se demonstreze ca :1 / (x^2 + y + 1) + 1 / (y^2 + z + 1) + 1 / (z^2 + x + 1) <= 1 Astept mesaj privat cu demonstratia completa. Primii trei vor primi un set de puncte, acum ma gandesc la un joc care sa va motiveze sa postati probleme de orice natura si sa va faca placere sa le rezolvati Edited January 5, 2013 by begood Quote Link to comment Share on other sites More sharing options...
daNNy.bv Posted January 5, 2013 Report Share Posted January 5, 2013 ce vrea sa insemne ^ ?puterea? Quote Link to comment Share on other sites More sharing options...
TheTime Posted January 5, 2013 Report Share Posted January 5, 2013 Pentru x = y = z = 1,x + y + x = xy + yz + xz = 3, este adevarata, dar1/3 + 1/3 + 1/3 = 1. Si 1 nu este mai mic decat 1... Voiai sa zici "<=" ? Quote Link to comment Share on other sites More sharing options...
begood Posted January 5, 2013 Author Report Share Posted January 5, 2013 da, multumesc TheTime. Editat si scuzati typo-ul. Quote Link to comment Share on other sites More sharing options...
daNNy.bv Posted January 5, 2013 Report Share Posted January 5, 2013 Eu am folosit metoda bruteforce : FAIL Quote Link to comment Share on other sites More sharing options...
begood Posted January 5, 2013 Author Report Share Posted January 5, 2013 ma uit maine ce ai scris acolo Quote Link to comment Share on other sites More sharing options...
fulminator Posted January 5, 2013 Report Share Posted January 5, 2013 Am dovedit prin inductie matematica, adica, inlocuind cu valori 1,2,3... x, y, z. Cum sunt > 0, e clar ca si puterile lor (2 si 3) sunt pozitive. Devine logic atunci ca pt orice valoare cu care inlocuiesti x, y, z, vei dovedi cerinta.Oricum, nu este solutia asteptata, desi e corecta, se numeste "matematica de mahala" si nu este acceptata in lucrari. Mai pe sleau, e logic, dar nu gasesc unde sa reduc. Quote Link to comment Share on other sites More sharing options...
SilvaDark Posted January 5, 2013 Report Share Posted January 5, 2013 @daNNy.bv: Ai gresit la linia 3 cand ai scos factor comun (y^2+z+1). Quote Link to comment Share on other sites More sharing options...
daNNy.bv Posted January 6, 2013 Report Share Posted January 6, 2013 care e gresala ca nu ma prind Quote Link to comment Share on other sites More sharing options...
SilvaDark Posted January 6, 2013 Report Share Posted January 6, 2013 (y^2+z+1)[(z^2+x+1)+(x^2+y+1)]+(x^2+y+1)(z^2+x+1)<..... Quote Link to comment Share on other sites More sharing options...
TheTime Posted January 6, 2013 Report Share Posted January 6, 2013 shaggi, e frumos codul, dar x,y,z pot fi numere reale. Quote Link to comment Share on other sites More sharing options...
daNNy.bv Posted January 7, 2013 Report Share Posted January 7, 2013 ai dreptate silver am omis aia Quote Link to comment Share on other sites More sharing options...
bcman Posted January 7, 2013 Report Share Posted January 7, 2013 Si eu am reusit sa rezolv pentru numere intregi (defapt naturale, pentru ca x,y,z > 0).Inmultim inecuatia cu x^2+y+1 si ajungem la 1 < x^2+y+1, adica x^2+y+1 > 1. Scadem un 1 si ramanem cu x^2 + y > 0. Aceasta propozitie matematica e adevarata pentru ca x si y sunt strict mai mari decat 0. Analog si pentru celelalte fractii. Astfel ajungem la o suma de 3 numere mai mici ca 1. In N si Z suma a trei numere mai mici ca 1 e si ea mai mica decat 1. In Q problema se complica. Doar atat am reusit sa fac. Quote Link to comment Share on other sites More sharing options...
Cril Posted January 7, 2013 Report Share Posted January 7, 2013 (edited) //link sters.Nu am fost prea atent, dar cred ca e ok facut Edited January 7, 2013 by Cril Quote Link to comment Share on other sites More sharing options...
TheTime Posted January 7, 2013 Report Share Posted January 7, 2013 Cril, nu e corect. Nu ai demonstrat nicaieri ca x^2 + y >=2, nici nu ai cum.Pentru x=1, y=1/2, z=2, ai x^2 + Y <2.Confunzi ceea ce trebuie sa demonstrezi cu ceea ce stii deja. Quote Link to comment Share on other sites More sharing options...
Cril Posted January 7, 2013 Report Share Posted January 7, 2013 (edited) Cril, nu e corect. Nu ai demonstrat nicaieri ca x^2 + y^2 >=2, nici nu ai cum.Pentru x=1, y=1/2, z=2, ai x^2 + Y^2 <2.e x^2+y >= 2.Am incercat sa sterg, dar vad ca la scanare se vede stersura.// mda, tot nu e bine(cred). Edited January 7, 2013 by Cril Quote Link to comment Share on other sites More sharing options...
begood Posted January 7, 2013 Author Report Share Posted January 7, 2013 hai, cine sparge gheata ? Quote Link to comment Share on other sites More sharing options...
ralex Posted January 7, 2013 Report Share Posted January 7, 2013 am spart gheata ) Quote Link to comment Share on other sites More sharing options...
begood Posted January 7, 2013 Author Report Share Posted January 7, 2013 ralex felicitari lu iubita ) Quote Link to comment Share on other sites More sharing options...
