设a>0,b>0,c>0,a≠b,b≠c,c≠a,且a,b,c满足a+b>c,求证：a^3+b^3+c^3+3abc>2(a+b)c^2

设a>0,b>0,c>0,a≠b,b≠c,c≠a,且a,b,c满足a+b>c,求证：a^3+b^3+c^3+3abc>2(a+b)c^2
设a>0,b>0,c>0,a≠b,b≠c,c≠a,且a,b,c满足a+b>c,求证：a^3+b^3+c^3+3abc>2(a+b)c^2

设a>0,b>0,c>0,a≠b,b≠c,c≠a,且a,b,c满足a+b>c,求证：a^3+b^3+c^3+3abc>2(a+b)c^2
a^3+b^3=（a+b）(a^2-ab+b^2)>c(a^2-ab+b^2),
所以a^3+b^3+c^3+3abc>c(a^2-ab+b^2)+c^3+3abc=a^2c+b^2c+c^3+2abc
因为a^2c+b^2c+c^3+2abc-2（a+b）c^2
=c[a^2+b^2+2ab+c^2-2(a+b)c]
=c[(a+b)^2+c^2-2(a+b)c]
=c(a+b-c)^2>0
所以a^2c+b^2c+c^3+2abc>2（a+b）c^2
所以a^3+b^3+c^3+3abc>2（a+b）c^2