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f:R->R,f(x)=x^2+3x+2. Sa se arate ca f(a)+f(a+1)>=0,oricare ar fi a€R

Răspuns :

f(a)=a^2+3a+2
f(a+1)=(a+1)^2+3(a+1)+2=a^2+2a+1+3a+3+2=a^2+5a+6

f(a)+f(a+1)>=0 <=> a^2+3a+2+a^2+5a+6>=0 <=> 2a^2+8a+8>=0 <=> a^2+4a+4>=0 <=> (a+2)^2>=0, ceea ce este adevarat, oricare ar fi a apartine R.