[LightLucifer]

ข้อความเดิมเขียนโดยคุณ nooonuii

6. $a,b,c\in\mathbb{R},$ pairwise distinct

$$\Big(\dfrac{a}{b-c}\Big)^2+\Big(\dfrac{b}{c-a}\Big)^2+\Big(\dfrac{c}{a-b}\Big)^2\geq 2$$

My Solution:

ให้ $a+b+c=S$
$\Big(\dfrac{a}{b-c}\Big)^2+\Big(\dfrac{b}{c-a}\Big)^2+\Big(\dfrac{c}{a-b}\Big)^2\geq 2$
$\leftrightarrow \Big(\dfrac{a+b+c-b-c}{b-c}\Big)^2+\Big(\dfrac{a+b+c-c-a}{c-a}\Big)^2+\Big(\dfrac{a+b+c-a-b}{a-b}\Big)^2\ge 2$
$\leftrightarrow \Big(\dfrac{S-b-c}{b-c}\Big)^2+\Big(\dfrac{S-c-a}{c-a}\Big)^2+\Big(\dfrac{S-a-b}{a-b}\Big)^2\ge 2$
ให้ $\dfrac{S-2b}{b-c}=x,\dfrac{S-2c}{c-a}=y,\dfrac{S-2a}{a-b}=z$
จะได้
$xyz=(x+2)(y+2)(z+2)\rightarrow -2(xy+yz+zx)=4(x+y+z)+8$
พิจรณา
$\Big(\dfrac{S-b-c}{b-c}\Big)^2+\Big(\dfrac{S-c-a}{c-a}\Big)^2+\Big(\dfrac{S-a-b}{a-b}\Big)^2=(x+1)^2+(y+1)^2+(z+1)^2$
$=(x+y+z)^2-2(xy+yz+zx)+2(x+y+z)+3=(x+y+z)^2+6(x+y+z)+11=(x+y+z+3)^2+2 \ge 2 \ \ \ \square$