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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/26 05:02:20
![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](/uploads/image/z/13362280-16-0.jpg?t=%E8%AE%BEa%3E0%2Cb%3E0%2Cc%3E0%2Ca%E2%89%A0b%2Cb%E2%89%A0c%2Cc%E2%89%A0a%2C%E4%B8%94a%2Cb%2Cc%E6%BB%A1%E8%B6%B3a%2Bb%3Ec%2C%E6%B1%82%E8%AF%81%EF%BC%9Aa%5E3%2Bb%5E3%2Bc%5E3%2B3abc%3E2%28a%2Bb%29c%5E2)
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