Cho x = by + cz (1); y = ax + cz (2); z = ax + by (3) và x + y + z ≠ 0; xyz ≠ 0.

Chứng minh đẳng thức $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$ .

x = by + cz (1); y = ax + cz (2); z = ax + by (3) và x +y + z ≠ 0; xyz ≠ 0.

Lấy (1) trừ (2), ta được:

x – y = (by + cz) − (ax + cz)

Û x – y = by – ax + cz – cz

Û x – y = by – ax

Û x + ax = by + y

Û x(a + 1) = y(b + 1) (*)

Lấy (2) trừ (3), ta được:

y – z = (ax + cz) − (ax + by)

Û y – z = ax – ax + cz – by

Û y – z = cz – by

Û y + by = z + cz

Û y(b + 1) = z (c + 1) (**)

Lấy (1) trừ (3), ta được:

x – z = (by + cz) − (ax + by)

Û x – z by – by + cz – ax

Û x – z = cz – ax

Û x + ax = cz + z

Û x(1 + a) = z(c + 1) (***)

Từ (*), (**), (***) suy ra: x(a + 1) = y(b + 1) = z(c + 1)

Đặt x(a + 1) = y(b + 1) = z(c + 1) = t

$\Rightarrow \left\{\begin{array}{c}a+1=\frac{t}{x}\\ b+1=\frac{t}{y}\\ c+1=\frac{t}{z}\end{array}\right.$ (do x, y, z ≠ 0)

Thay vào biểu thức $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}$ , ta được:

$$\begin{gathered} \frac{1}{\frac{t}{x}}+\frac{1}{\frac{t}{y}}+\frac{1}{\frac{t}{z}}=\frac{x}{t}+\frac{y}{t}+\frac{z}{t}=\frac{x+y+z}{t} \\ \Rightarrow \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{x+y+z}{t} \end{gathered}$$

Với x = by + cz (1); y = ax + cz (2); z = ax + by (3)

t = x(a+1) = ax + x = ax + by + cz

Ta có: $\frac{x+y+z}{t}=\frac{by+cz+ax+cz+ax+by}{ax+by+cz}$

$$\begin{gathered} =\frac{ax+ax+by+by+cz+cz}{ax+by+cz} \\ =\frac{2ax+2by+2cz}{ax+by+cz} \\ =\frac{2\left(ax+by+cz\right)}{ax+by+cz}=2 \end{gathered}$$

Vậy $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$ .