Prove the inequality $\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74$ Inequality

Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove the following inequality
  $$\frac 1a + \frac 1b +\frac 1c \ge \frac{a^3+b^3+c^3}{3} +\frac 74.$$

I stumbled upon this question some days ago and been trying AM-GM to find the solution but so far have been unsuccessful.

here is a elementary method:
WLOG, let $a=$Min{$a,b,c$} $\implies a \le1 ,x=bc$
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=3((a+b+c)^2-3ab-3ac-3bc)=3(9-3a(3-a)-3bc)$
LHS$=\dfrac{a(3-a)+bc}{abc}=\dfrac{a(3-a)+x}{ax},$ 
RHS$= 9-3a(3-a)-3bc+abc+\dfrac{7}{4}=\dfrac{43}{4}-3a(3-a)+(a-3)x$ 
LHS-RHS$=\dfrac{(3-a)}{x}\left(x^2-\dfrac{4a^3-36a^2+43a-4}{4a(3-a)}x+1 \right)$
$f(x)=x^2+\dfrac{g_1(a)}{4a(3-a)}x+1 ,g_1(a)=-4a^3+36a^2-43a+4$
it is trivial that $g_1(a)>0 \implies f(x)>0 $
when $g_1(a)<0 ,f_{min}=1-\left(\dfrac{g_1(a)}{2\times4a(3-a)}\right)^2=\dfrac{(2a-1)^2(4-3a)(12a^3-44a^2+67a-4)}{64a^2(3-a)^2}$
$12a^3-44a^2+67a-4+g_1(a)=8a(3-a)>0 \implies 12a^3-44a^2+67a-4>0 \implies \\f_{min} \ge 0$
when $a=\dfrac{1}{2}, f_{min}=0 \implies x= -\dfrac{g_1(a)}{2\times4a(3-a)}=1$
$b+c=\dfrac{5}{2},bc=1 \implies (\dfrac{1}{2},2)$
