Question

设a>0b>0c>0求证a^2/b+b^2/c+c^2/a≥a+b+c

Answer 1

a^2/b+b^2/c+c^2/a=a*a/b+b*b/c+c*c/a

由契贝谢夫不等式：

【a*a/b+b*b/c+c*c/a】/3>=【a+b+c】/3*【a/b+b/c+c/a】/3

又a/b*b/c*c/a=1 有a/b+b/c+c/a>=3.

【a*a/b+b*b/c+c*c/a】/3>=>【a+b+c】/3*【a/b+b/c+c/a】/3>=【a+b+c】/3

所以a^2/b+b^2/c+c^2/a》=a+b+c.