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.
 A: 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 $$(a-1)^3+(b-1)^3+(c-1)^3\geq {3\over 4}(a+b+c)-3= -{3\over 4}$$
A: 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.
A: 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.
A: The homogenisation gives
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$
which is true by Schur.
Indeed, we need to prove that
$$\sum_{cyc}(a-1)^3+\frac{3}{4}\geq0$$ or
$$\sum_{cyc}\left(a^3-3a^2+3-1+\frac{1}{4}\right)\geq0$$ or
$$\sum_{cyc}(a^3-(a+b+c)a^2)+\frac{27}{4}$$ or
$$4\sum_{cyc}(-a^2b-a^2c)+(a+b+c)^3\geq0$$ or
$$\sum_{cyc}(-4a^2b-4a^2c+a^3+3a^2b+3a^2c+2abc)\geq0$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which is obvious.
A: 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 $$-\frac{3}{4}\left(2a-3\right)\left(4a^2-6a+3\right)\ge 0$$
