Let a, b and c be positive real numbers satisfying Let a, b and c be positive real numbers satisfying 
$\frac{1}{a+2019}$ + $\frac{1}{b+2019}$ +$\frac{1}{c+2019}$ = $\frac{1}{2019}$
Show that abc ≥$4038^3$.
My first impression is to use arithematic mean ≥ geometric mean.
 A: Hint: Your equation can be written as $$4076361(a+b+c)+16460345718-abc=0$$
Substituting $$x=2019$$ you will get $$2x^3+x^2(a+b+c)-abc=0$$
So we have $$abc=2x^3+x^2(a+b+c)$$ and
$$\frac{abc-2x^3}{x^2}=a+b+c\geq 3\sqrt[3]{abc}$$ and from here we get
$$(abc-2x^3)^3\geq 27x^6abc$$
Expanding
$$(abc)^3-6x^3(abc)^2-15abcx^6-8x^9\geq 0$$
with $$abc=t$$ we get the function
$$f(t)=t^3-6t^2x^3-15tx^6-8x^9$$
So now we must compute the positive zero of $$f(t)$$
and we get $$t=abc\geq 65841382872=4038^3$$ and $$f(t)\geq 0$$ is fulfilled.
A: With the help of the Lagrange multipliers the question can be stated as:
determine the stationary points for
$$
L(a,b,c,\lambda) = a b c - (2N)^3 + \lambda\left(\frac{1}{a+N}+\frac{1}{b+N}+\frac{1}{c+N}-\frac{1}{N}\right)
$$
giving the stationary conditions
$$
\left\{
\begin{array}{rcl}
 b c-\frac{\lambda }{(a+N)^2}=0 \\
 a c-\frac{\lambda }{(b+N)^2}=0 \\
 a b-\frac{\lambda }{(c+N)^2}=0 \\
 \frac{1}{a+N}+\frac{1}{b+N}+\frac{1}{c+N}-\frac{1}{N}=0 \\
\end{array}
\right.
$$
easily solved given the feasible values
$$
a = b = c = 2N, \lambda = 36N^4
$$
then 
$$
abc \ge (2N)^3
$$
because the solution point is a minimum point.
NOTE
Substituting $c = \frac{N^2(a+b+2N)}{N^2-ab}$ into $f(a,b,c) = a b c - (2N)^3$ we get
$$
g(a,b) = N^2 \left(\frac{a b (a+b+2 N)}{a b-N^2}-8 N\right)
$$
and the hessian gives
$$
H_g = \left(
\begin{array}{cc}
 -\frac{2 b N^4 (b+N)^2}{\left(N^2-a b\right)^3} & -\frac{2 N^5 (a+N) (b+N)}{\left(N^2-a b\right)^3} \\
 -\frac{2 N^5 (a+N) (b+N)}{\left(N^2-a b\right)^3} & -\frac{2 a N^4 (a+N)^2}{\left(N^2-a b\right)^3} \\
\end{array}
\right)
$$
which at the solution point has the eigenvalues $\{2N, \frac{2N}{3}\}$ hence $H_g$ is definite positive, characterizing the minimum.
