Tìm a,b,c thoả mãn a.b.c

Ta có:
1/a + 1/b + 1/c = (a+b)/c + (b+c)/a + (c+a)/b = 1/3

Từ đó, ta có:
(a+b+c)(1/a + 1/b + 1/c) = 1/3(a+b+c)

(a+b+c)(1/a + 1/b + 1/c) = (a+b+c)/3

(a+b+c)(1/a + 1/b + 1/c) - (a+b+c)/3 = 0

(a+b+c)(1/a + 1/b + 1/c - 1/3) = 0

(a+b+c)(3(ab+bc+ca) - (a+b+c))/3abc = 0

Vì a.b.c khác 0, nên ta có:
3(ab+bc+ca) - (a+b+c) = 0

3ab + 3bc + 3ca - a - b - c = 0

2ab + 2bc + 2ca = a + b + c

2(ab + bc + ca) = a + b + c

2(ab + bc + ca) = 2020

(a+b+c)(ab+bc+ca) = 2020

(a+b+c)(ab+bc+ca) = (a+b+c)^2 - (a^2 + b^2 + c^2)

(a+b+c)^2 - (a^2 + b^2 + c^2) = 2020

a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - a^2 - b^2 - c^2 = 2020

2ab + 2bc + 2ca = 2020

Vậy a+b+c = 2020.