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.

You can prove that $$\color{red}{(x-1)^3\geq {3\over 4}x-1}$$ for all nonegative $x$; it is equivalent to $x(2x-3)^2\ge 0$, so $$(a-1)^3+(b-1)^3+(c-1)^3\geq {3\over 4}(a+b+c)-3= -{3\over 4}$$
