Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$ 
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
  $$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$

Here's what I did.
Let $c \ge a \ge b$.
We have that 
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a - 2)(c^2 + a^2 - ca - c - a + 1)\\
&\ge (1 - b)\left[\dfrac{3}{4}(c + a)^2 - b + 4\right]\\
&=(1 - b)\left[\dfrac{3}{4}(3 - b)^2 - b + 4\right]\\ 
&= \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43)
\end{align*}
That means 
\begin{align*}
(a - 1)^3 + (b - 1)^3 + (c - 1)^3 &\ge \dfrac{1}{4}(1 - b)(3b^2 - 22b + 43) + (b - 1)^3\\
&= \dfrac{1}{4}(b - 1)(b^2 + 14b - 39)
\end{align*}
Since $c \ge a \ge b \implies b \le \dfrac{a + b + c}{3} = 1$.
And this is where I am stuck right now.

My sketchy proof:
First you need to prove that if $x+y+z=0$, then $x^3 + y^3 + z^3 = 3xyz$
Since $a+b+c=3$, we have $(a - 1)+(b-1)+(c-1)=0$, therefore, $S = (a - 1)^3+(b-1)^3+(c-1)^3=3(a - 1)(b-1)(c-1)$
Replace $c = 3 - a -b$, we have $S = 3(a-1)(b-1)(2-a-b)$
Expand this, we have $S/3 = a^2 + b^2 + 4ab - 3(a+b) - ab(a+b) + 2$
We can safely assume that $a \geq b$, since the order $a$, $b$ doesn't matter. So we have: $(a-b)^2 \geq 0$ or $a^2 + b^2 \geq 2ab$. Equality happens when $a = b$.
So $S/3 \geq 6ab - 3(a+b) - ab(a+b) + 2$
Since $c \geq 0$, we have $3 - a - b \geq 0$, or $-(a+b) \geq -3$. Equality happens when $a + b = 3$.
So $S/3 \geq 6ab + 3(-3) + ab(-3) + 2 = 3ab - 7$, or $S \geq 9ab - 21$. Equality happens when $a = b$ and $a + b = 3$, or $a = b = 3/2$. Replace $a$, $b$ in $9ab-21$ with $a = b = 3/2$ we have $S \geq -3/4$
I'm not satisfy with the last step, though.
