Question

fie a,b, c numere reale a.î a patrat + b patrat + c patrat + ab+ bc + ca =6 .aratati ca | a+b+c| mai mic sau egal cu 3.

Answer 1

[tex]Din \ (a-b)^2+(a-c)^2+(b-c)^2 \geq 0=\ \textgreater \ a^2+b^2+c^2 \geq ab+ac+bc\\ 6-(ab+ac+bc) \geq ab+ac+bc=\ \textgreater \ ab+ac+bc \leq 3(1)\\ |a+b+c|= \sqrt{(a+b+c)^2} = \sqrt{a^2+b^2+c^2+2ab+2ac+2bc}=\\ = \sqrt{6+ab+ac+bc} \leq 3(adevarat)\ deoarece\\ \sqrt{6+ab+ac+bc} \leq 3\ \textless \ =\ \textgreater \ \sqrt{6+ab+ac+bc} ^2 \leq 3^2\ \textless \ =\ \textgreater \ \\ 6+ab+ac+bc \leq 9\ \textless \ =\ \textgreater \ ab+ac+bc \leq 3(1) [/tex]