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.

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$\frac{3v^2}{w^3}\geq\frac{9u^3-9uv^2+w^3}{u^4}+\frac{7}{4u},$$
which is a linear inequality of $v^2$, 
which says that it's enough to prove our inequality for an extremal value of $v^2$, 
which happens for equality case of two variables.
Let $b=a$.
Hence, $c=3-2a$, where $0<a<\frac{3}{2}$ and we get
$$(2a-1)^2(24-45a+26a^2-4a^3)\geq0,$$
which is true for $a\in\left(0,\frac{3}{2}\right)$.
Done!
