<s>[INST] 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.
 [/INST] Perhaps a different approach than you are using. Using the Lagrange multipliers, we define the auxiliary problem
$$ F = (x-1)^3 + (y-1)^3 + (z-1)^3 - \lambda(x + y + z - 3). $$
Differentiating and setting to $0$ we obtain the conditions
\begin{align}
3(x-1)^2 = \lambda \\
3(y-1)^2 = \lambda \\
3(z-1)^2 = \lambda
\end{align}
It is easy to see that if we set $x,y,z$ to the critical point obtained by these conditions we will reach a contradiction with the constraint $x + y + z = 3$ (check this yourself). Then, it must be that the extremum if it exists must have at least one variable (say $z$) which is $0$. This leads to a value of $\lambda = 3/4$ and finally, one can show that any permutation of the triplet $(3/2,3/2,0)$ leads to the minimum value. You should also check that this is indeed the minimum.
</s>