#(2x)/(2x^2+5x+2) > 1/(x+1)# where #x!=-1,-1/2,-2#
#(2x)/(2x^2+5x+2) -1/(x+1) >0#
#(2x(x+1)-(2x^2+5x+2))/((x+1)(2x^2+5x+2)) >0#
#(2x^2+2x-2x^2-5x-2)/((x+1)(2x^2+5x+2)) >0#
#(-3x-2)/((x+1)(2x^2+5x+2)) >0#
#((x+1)(2x^2+5x+2))^2times(-3x-2)/((x+1)(2x^2+5x+2)) >0times((x+1)(2x^2+5x+2))^2#
#color (red) ("you need to multiply both sides by the denominator squared because you don't know whether your denominator is a positive or negative number. If it was negative, then your inequality sign would change."#
#(-3x-2)(x+1)(2x^2+5x+2) >0#
#(-3x-2)(x+1)(2x+1)(x+2)>0#
#-(3x+2)(x+1)(2x+1)(x+2)>0#
graph{(-3x-2)(x+1)(2x+1)(x+2) [-10, 10, -5, 5]}
looking at the graph, we can see that
#-2/3 < x < -1/2# and #-2 < x <-1# only
