[tex]\displaystyle f(x)=2x^3-6x+1 \\ \\ f'(x)=6x^2-6=6(x-1)(x+1). \\ \\ Pentru~x \in [1, + \infty)~avem~f'(x) \ge 0,~deci~functia~f~este~crescatoare \\ \\ pe~intervalul~[1, + \infty). \\ \\ Rezulta~f(2012) \le f(2013)~si~f(2014) \le f(2015).~De~aici~rezulta \\ \\ \boxed{f(2012)+f(2014) \le f(2013)+f(2015).}[/tex]
