Question

Aflati valoarea minima a expresiei:
E(x,y)= | xy - 2 x - y +2| + | -xy - x +y +1 | + | 3x+3| , x,y sunt reale.
multumesc anticipat :)

Answer 1

$Vom~folosi~inegalitatea~|m|+|n| \geq |m+n|~cu~egalitatea~daca: \\ \\ -cel~putin~unul~dintre~numerele~m~si~n~este~egal~cu~0 \\ \\ -ambele~numere~sunt~de~acelasi~semn \\ \\ Asadar:~|xy-2x-y+2|+|-xy-x+y+1| \geq \\ \\ \geq |(xy-2x-y+2)+(-xy-x+y+1)|=|-3x+3|. \\ \\ E(x,y) \geq |-3x+3|+|3x+3| \geq |(-3x+3)+(3x+3)|=|6|=6. \\ \\ E(x,y) \geq 6 \Rightarrow \boxed{min ~E(x,y)=6}~.$

$Valoarea~minima~a~lui~E(x,y)~este~6.~ramane~sa~verificam \\ \\ daca~aceasta~poate~avea~loc. \\ \\ Observam~ca~E(0,0)=6,~deci~minimul~poate~fi~atins.$