如何证明（2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc>0

a>b>c>0
则2a-b-c>2b-a-c>2c-a-b
lna>lnb>lnc
由排序不等式
同序和>乱序和>逆序和
(2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc>(2b-a-c)lna+(2a-b-c)lnb+(2a-b-c)lnc
(2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc>(2c-a-b)lna+(2c-a-b)lnb+(2b-a-c)lnc
(2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc=(2a-b-c)lna+(2b-a-c)lnb+(2c-a-c)lnc
三式相加得
3[(2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc]>0*lna+0*lnb+0*lnc
即(2a-b-c)lna+(2b-a-c)lnb+(2c-a-b)lnc>0