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Ta có:$a + b + 1 = x^3 + y^3 + 1 = (x + y)(x^2 - xy + y^2) + 1$

Đặt $a = x^3, b = y^3, c = z^3 \implies xyz = 1$.

Ta có:$a + b + 1 = x^3 + y^3 + 1 = (x + y)(x^2 - xy + y^2) + 1$

Vì $x^2 - xy + y^2 \ge xy$ với mọi $x, y > 0$, nên:$a + b + 1 \ge (x + y)xy + 1 = x^2y + xy^2 + xyz = xy(x + y + z)$

Lấy nghịch đảo hai vế (đổi chiều BĐT):

$$\frac{1}{a+b+1} \le \frac{1}{x^2y + xy^2 + xyz} = \frac{z}{xyz(x+y+z)} = \frac{z}{x+y+z}$$

$$\frac{1}{b+c+1} \le \frac{1}{y^2z + yz^2 + xyz} = \frac{x}{xyz(x+y+z)} = \frac{x}{x+y+z}$$

$$\frac{1}{c+a+1} \le \frac{1}{z^2x + zx^2 + xyz} = \frac{y}{xyz(x+y+z)} = \frac{y}{x+y+z}$$

$$=>\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} \le \frac{z+x+y}{x+y+z} = 1$$