Question

Chứng minh rằng: 
\(\left(\dfrac{a+b}{a-b}\right)^2+\left(\dfrac{b+c}{b-c}\right)^2+\left(\dfrac{c+a}{c-a}\right)^2\ge2\)

Answer 1

Lời giải:
Đặt $\frac{a+b}{a-b}=x; \frac{b+c}{b-c}=y; \frac{c+a}{c-a}=z$. Khi đó:

$xy+yz+xz=\frac{(a+b)(b+c)}{(a-b)(b-c)}+\frac{(a+b)(c+a)}{(a-b)(c-a)}+\frac{(b+c)(c+a)}{(b-c)(c-a)}$

$=\frac{(a+b)(b+c)(c-a)+(b+c)(c+a)(a-b)+(c+a)(a+b)(b-c)}{(a-b)(b-c)(c-a)}=-1$
Suy ra:

$(\frac{a+b}{a-b})^2+(\frac{b+c}{b-c})^2+(\frac{c+a}{c-a})^2=x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$

$=(x+y+z)^2+2\geq 2$

Ta có đpcm.