Câu hỏi của nguyễn thị thảo vân

Question

cho ba số a,b,c đôi một khác nhau: CM:

$\frac{\left(a+b\right)^2}{\left(a-b\right)^2}+\frac{\left(b+c\right)^2}{\left(b-c\right)^2}+\frac{\left(c+a\right)^2}{\left(c-a\right)^2}\ge2$

giải chi tiết giúp mk nha cảm ơn nhiều

Trần Đức Thắng · Accepted Answer

Đặt $x=\frac{a+b}{a-b};y=\frac{b+c}{b-c};z=\frac{c+a}{c-a}$Ta có : $x+1=\frac{2a}{a-b};y+1=\frac{2b}{b-c};z+1=\frac{2c}{c-a}$ (1)$x-1=\frac{2b}{a-b};y-1=\frac{2c}{b-c};z-1=\frac{2a}{c-a}$ (2)Từ (1) và (2) => $\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(x-1\right)\left(y-1\right)\left(z-1\right)$<=> $\left(xy+x+y+1\right)\left(z+1\right)=\left(xy-x-y+1\right)\left(z-1\right)$<=> $xyz+xz+yz+z+xy+x+y+1=xyz-xz-yz+z-xy+x+y-1$<=> $xy+yz+xz=-1$TA có $\left(x+y+z\right)^2\ge0\Leftrightarrow x^2+y^2+z^2\ge-2\left(xy+yz+xz\right)=2$Câu trả lời: Đặt $x=\frac{a+b}{a-b};y=\frac{b+c}{b-c};z=\frac{c+a}{c-a}$Ta có : $x+1=\frac{2a}{a-b};y+1=\frac{2b}{b-c};z+1=\frac{2c}{c-a}$ (1)$x-1=\frac{2b}{a-b};y-1=\frac{2c}{b-c};z-1=\frac{2a}{c-a}$ (2)Từ (1) và (2) => $\left(x+1\right)\left(y+1\right)\left(z+1\right)=\left(x-1\right)\left(y-1\right)\left(z-1\right)$<=> $\left(xy+x+y+1\right)\left(z+1\right)=\left(xy-x-y+1\right)\left(z-1\right)$<=> $xyz+xz+yz+z+xy+x+y+1=xyz-xz-yz+z-xy+x+y-1$<=> $xy+yz+xz=-1$TA có $\left(x+y+z\right)^2\ge0\Leftrightarrow x^2+y^2+z^2\ge-2\left(xy+yz+xz\right)=2$