Question

Fie numerele reale a,b, c care apartin intervalului [0,2] astfel incat ab+bc+ca=2. Demonstrati ca a^2 + b^2 + c^2 <= 6.
Multumesc!

Answer 1

$\displaystyle Asa~cum~am~spus~si~in~comentariu,~desi~in~enutul~original~apare \\ \\ " \leq 6",~corect~este~" \leq 5".~(ma~rog,~e~doar~o~intarire) \\ \\ ---------- \\ \\ Avem~de~demonstrat~ca~a^2+b^2+c^2 \leq 5 \Leftrightarrow \\ \\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac \leq 5+2ab+2bc+2ac \Leftrightarrow \\ \\ \Leftrightarrow (a+b+c)^2 \leq 9 \Leftrightarrow a+b+c \leq 3.$

$\displaystyle Presupunem~prin~reducere~la~absurd~ca~a+b+c\ \textgreater \ 3. \\ \\ Din~enunt~avem~(a-2)(b-2)(c-2) \leq 0 \Leftrightarrow \\ \\ \Leftrightarrow abc-2(ab+bc+ac)+4(a+b+c)-8 \leq 0 \Leftrightarrow \\ \\ \Leftrightarrow abc+4(a+b+c) \leq 12.~(*) \\ \\ Avem~a,b,c \geq 0,~deci~abc \geq 0,~si~tinand~cont~de~presupunerea \\ \\ facuta,~avem~abc+4(a+b+c)\ \textgreater \ 0+4 \cdot 3=12,~contradictie~cu ~(*). \\ \\ Prin~urmare~presupunerea~facuta~a~fost~falsa,~ceea~ce~inseamna \\ \\ ca~a+b+c \leq 3,~q.e.d.$