Đặt $x=b+c-a$
$y=a+c-b$
$z=a+b-c$
$⇒x+z=b+c-a+a+b-c=2b$
$x+y=b+c-a+a+c-b=2c$
$y+z=a+c-b+a+b-c=2a$
Ta có: `(y+z)/x+(x+z)/y+(x+y)/z`
`=y/x+z/x+x/y+z/y+x/z+y/z`
`=(y/x+x/y)+(z/y+y/z)+(x/z+z/x)`
Dễ chứng minh `a/b+b/a≥2` với $a,b>0$
`⇒(y/x+x/y)+(z/y+y/z)+(x/z+z/x)≥6`
`⇒(y+z)/x+(x+z)/y+(x+y)/z≥6`
`⇒(2a)/(b+c-a)+(2b)/(a+c-b)+(2c)/(a+b-c)≥6`
`⇒(a)/(b+c-a)+(b)/(a+c-b)+(c)/(a+b-c)≥3` $(Đpcm)$.