Question

一道不等式证明

已知a>b>c,求证a2/(a-b)+b2/(b-c)>a+2b+c

Answer 1

a2/(a-b)+(a-b)>2a(基本不等式)
b2/(b-c)+（b-c)>2b
所以a2/(a-b)+(a-b)+b2/(b-c)+（b-c)>2a+2b
所以a2/(a-b)+b2/(b-c)>a+2b+c

Answer 2

令a-b=x>0，b-c=y>0,则a=c+x+y, b=c+y
a^2/(a-b)+b^2/(b-c)
=(c+x+y)^2/x+(c+y)^2/y
=(c^2+x^2+y^2+2cx+2cy+2xy)/x + (c^2+y^2+2cy)/y
=(c^2+y^2+2cy)/x +x+2c+2y +(c^2)/y +y+2c
=(c+y)^2/x+(c^2)/y+(4c+3y+x)
=b^2/x+c^2/y +(a+2b+c)>a+2b+c