If a=b=c≠0, the expression evaluates as ½+½+½=3/2. So the expression is not >3/2, and the inequality is untrue. The smallest values for (a,b,c)=(1,2,3) if we reject zero as an integer, and the inequality is true: ⅕+½+1=17/10>3/2. If we accept zero as an integer, then (a,b,c)=(0,1,2) and we get 0+½+2=5/2>3/2, so the inequality is true. For small integers the inequality is true and the sum of the fractions exceeds 3/2.
Because of symmetry we only need to consider a<b<c. When a becomes very large, and b and c are close to a, the inequality remains true but the sum of the fractions gets closer to 3/2 and is always larger. For example, if b=a+1 and c=a+2, the sum becomes a/(2a+3)+½+(a+2)/(2a+1)=½+(a(2a+1)+(a+2)(2a+3))/(4a²+5a+3).
Expanding this we get:
½+(4a²+8a+6)/(4a²+5a+3)=½+(4a²+5a+3+3(a+1))/(4a²+5a+3), so the fraction is always bigger than 1 so the sum>3/2.