均值不等式问题设a,b,c都是正数,求证:1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/18 19:52:08
![均值不等式问题设a,b,c都是正数,求证:1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)](/uploads/image/z/5128754-50-4.jpg?t=%E5%9D%87%E5%80%BC%E4%B8%8D%E7%AD%89%E5%BC%8F%E9%97%AE%E9%A2%98%E8%AE%BEa%2Cb%2Cc%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%2C%E6%B1%82%E8%AF%81%EF%BC%9A1%2F2a%2B1%2F2b%2B1%2F2c%3E%3D1%2F%28b%2Bc%29%2B1%2F%28c%2Ba%29%2B1%2F%28a%2Bb%29)
均值不等式问题设a,b,c都是正数,求证:1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)
均值不等式问题
设a,b,c都是正数,求证:1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)
均值不等式问题设a,b,c都是正数,求证:1/2a+1/2b+1/2c>=1/(b+c)+1/(c+a)+1/(a+b)
1/4a+1/4b
=(a+b)/4ab
≥(a+b)/(a+b)^2
=1/(a+b)
同理1/4b+1/4c≥1/(b+c)
1/4c+1/4a≥1/(c+a)
由以上三式相加可得1/2a+1/2b+1/2c≥1/(a+b)+1/(b+c)+1/(c+a)
不等式左边可变为1/4a+1/4a+1/4b+1/4b+1/4c+1/4c
即:(1/4a+1/4b)+(1/4b+1/4c)+(1/4a+1/4c);
因为1/4a+1/4b=(a+b)/4ab,并由(a+b)^2≥4ab
得:1/4a+1/4b=(a+b)/4ab≥(a+b)/(a+b)^2=1/(a+b);
同理可得
...
全部展开
不等式左边可变为1/4a+1/4a+1/4b+1/4b+1/4c+1/4c
即:(1/4a+1/4b)+(1/4b+1/4c)+(1/4a+1/4c);
因为1/4a+1/4b=(a+b)/4ab,并由(a+b)^2≥4ab
得:1/4a+1/4b=(a+b)/4ab≥(a+b)/(a+b)^2=1/(a+b);
同理可得
1/4b+1/4c≥(b+c)/(b+c)^2=1/(b+c);
1/4a+1/4c≥(a+c)/(a+c)^2=1/(a+c);
最后将上面三式相加
得1/2a+1/2b+1/2c≥1/(a+b)+1/(b+c)+1/(c+a)
原题得证。
收起