Dtjytyk

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 + 4right]\
&=(1 - b)left[dfrac{3}{4}(3 - b)^2 - b + 4right]\
&= 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 $$color{red}{(x-1)^3geq {3over 4}x-1}$$ for all nonegative $x$; it is equivalent to $x(2x-3)^2ge 0$, so $$(a-1)^3+(b-1)^3+(c-1)^3geq {3over 4}(a+b+c)-3= -{3over 4}$$

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.

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.

Hope it will help!

WLOG $ bge age cRightarrow cle 1Rightarrow 1-cge 0$

We have: $$fleft(a,b,cright)=left(a-1right)^3+left(b-1right)^3+left(c-1right)^3$$

And $$fleft(frac{a+b}{2};frac{a+b}{2};cright)=2left(frac{a+b}{2}-1right)^3+left(c-1right)^3$$

Note that:$$fleft(a,b,cright)-fleft(frac{a+b}{2};frac{a+b}{2};cright)=frac{3}{4}left(a-bright)^2left(a+b-2right)=frac{3}{4}left(a-bright)^2left(1-cright)ge 0$$

So you need to prove this inequality in the case $a=b$ the condition gives $c=3-2a$

Thus we need to prove $$2left(a-1right)^3+left(3-2a-1right)^3ge -frac{3}{4}$$

Or $$-frac{3}{4}left(2a-3right)left(4a^2-6a+3right)ge 0$$

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
$$4sum_{cyc}(-a^2b-a^2c)+(a+b+c)^3geq0$$ 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.