<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] Hope it will help!
WLOG $ b\ge a\ge  c\Rightarrow c\le 1\Rightarrow 1-c\ge 0$
We have: $$f\left(a,b,c\right)=\left(a-1\right)^3+\left(b-1\right)^3+\left(c-1\right)^3$$
And $$f\left(\frac{a+b}{2};\frac{a+b}{2};c\right)=2\left(\frac{a+b}{2}-1\right)^3+\left(c-1\right)^3$$
Note that:$$f\left(a,b,c\right)-f\left(\frac{a+b}{2};\frac{a+b}{2};c\right)=\frac{3}{4}\left(a-b\right)^2\left(a+b-2\right)=\frac{3}{4}\left(a-b\right)^2\left(1-c\right)\ge 0$$
So you need to prove this inequality in the case $a=b$ the condition gives $c=3-2a$
Thus we need to prove $$2\left(a-1\right)^3+\left(3-2a-1\right)^3\ge -\frac{3}{4}$$
Or $$-\frac{3}{4}\left(2a-3\right)\left(4a^2-6a+3\right)\ge 0$$
</s>