Question

求证(2a-b/a-b)^2+(2b-c/b-c)^2+(2c-a/c-a)^2>=5

Answer 1

定理
设实数x,y,z满足
yz+zx+xy=x+y+z+t,则有
(x-k)^2+(y-k)^2+(z-k)^2≥2k^2-2k-1-2t
(1)
定理证明
(1)式展开为
k^2-2k(x+y+z-1)+x^2+y^2+z^2+2t+1≥0
<==>
k^2-2k(x+y+z-1)+(x+y+z)^2-2(x+y+z)+1≥0
<==>
k^2-2k(x+y+z-1)+(x+y+z-1)^2≥0
<==>
[k+1-(x+y+z)]^2≥0.
下面运用定理来证明不等式.
设a,b,c是互不相同的实数.试证
[(2a-b)/(a-b)]^2+[(2b-c)/(b-c)]^2+[(2c-a)/(c-a)]^2≥5
(2)
(2)<==>
[a/(a-b)+1]^2+[b/(b-c)+1]^2+[c/(c-a)+1]^2≥5
令x=a/(a-b),
y=b/(b-c),
z=c/(c-a).
易验证
yz+zx+xy+1=x+y+z.
故定理中取k=-1,t=-1,得
(x+1)^2+(y+1)^2+(z+1)^2≥2(-1)^2+2-1+2=5。
不等式（2）得证。