<s>[INST] 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.
 [/INST] 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.
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