[tex]\displaystyle 1.~Avem~E=xy-2x-2y+6=xy-2x-2y+4+2= \\ \\ =(x-2)(y-2)+2. \\ \\ Cum~x,y \ge 2 \Rightarrow x-2 \ge 0~si~y-2 \ge 0.~Deci~(x-2)(y-2) \ge 0. \\ \\ Rezulta~E \ge 2,~adica~E \in [2,+\infty).[/tex]
[tex]\displaystyle 2. ~Avem~de~demonstrat~ca~-1\ \textless \ \frac{x+y}{1+xy}\ \textless \ 1,~unde~x,y \in (-1,1). \\ \\ Cum~x,y \in (-1,1),~rezulta~ca~xy \in (-1,1),~deci~1+xy \in (0,2). \\ \\ Ne~intereseaza~doar~faptul~ca~1+xy\ \textgreater \ 0.~Aceasta~ne~permite \\ \\ sa~echivalam~inegalitatea~data~prin~inmultire~in~toti~membrii~cu~ \\ \\ (1+xy). \\ \\ Deci~trebuie~sa~demonstram~ca~-1-xy\ \textless \ x+y\ \textless \ 1+xy. [/tex]
[tex]\displaystyle \bullet~ Demonstram~ca~-1-xy\ \textless \ x+y: \\ \\ -1-xy\ \textless \ x+y \Leftrightarrow 0\ \textless \ 1+x+y+xy \Leftrightarrow 0\ \textless \ (x+1)(y+1), \\ \\ adevarat,~caci~x+1\ \textgreater \ 0~si~y+1\ \textgreater \ 0. \\ \\ \bullet ~Demonstram~ca~x+y\ \textless \ 1+xy: \\ \\ x+y\ \textless \ 1+xy \Leftrightarrow 0\ \textless \ 1-x-y+xy \Leftrightarrow 0\ \textless \ (1-x)(1-y), \\ \\ adevarat,~caci~1-x\ \textless \ 0~si~1-y\ \textless \ 0.[/tex]
