a、b、c为正实数,求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具体解释啊``]解:由均值不等式,得(a+b)^2+(a+b+4c)^2=(a+b)^2+[(a+2c)+(b+2c)]^2>=(2根ab)^2+[2(根(2ab))+2(根(2bc))]^2=4ab+8ac+8bc+16c根(ab)于是,[(a+b)^2+(a+
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/17 17:06:55
![a、b、c为正实数,求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具体解释啊``]解:由均值不等式,得(a+b)^2+(a+b+4c)^2=(a+b)^2+[(a+2c)+(b+2c)]^2>=(2根ab)^2+[2(根(2ab))+2(根(2bc))]^2=4ab+8ac+8bc+16c根(ab)于是,[(a+b)^2+(a+](/uploads/image/z/983536-16-6.jpg?t=a%E3%80%81b%E3%80%81c%E4%B8%BA%E6%AD%A3%E5%AE%9E%E6%95%B0%2C%E6%B1%82%5B%28a%2Bb%29%5E2%2B%28a%2Bb%2B4c%29%5E2%5D%28a%2Bb%2Bc%29%2Fabc%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC.+%5B%E6%B1%82%E9%AB%98%E6%89%8B%E5%85%B7%E4%BD%93%E8%A7%A3%E9%87%8A%E5%95%8A%60%60%5D%E8%A7%A3%3A%E7%94%B1%E5%9D%87%E5%80%BC%E4%B8%8D%E7%AD%89%E5%BC%8F%2C%E5%BE%97%28a%2Bb%29%5E2%2B%28a%2Bb%2B4c%29%5E2%3D%28a%2Bb%29%5E2%2B%5B%28a%2B2c%29%2B%28b%2B2c%29%5D%5E2%3E%3D%282%E6%A0%B9ab%29%5E2%2B%5B2%28%E6%A0%B9%282ab%29%29%2B2%28%E6%A0%B9%282bc%29%29%5D%5E2%3D4ab%2B8ac%2B8bc%2B16c%E6%A0%B9%28ab%29%E4%BA%8E%E6%98%AF%2C%5B%28a%2Bb%29%5E2%2B%28a%2B)
a、b、c为正实数,求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具体解释啊``]解:由均值不等式,得(a+b)^2+(a+b+4c)^2=(a+b)^2+[(a+2c)+(b+2c)]^2>=(2根ab)^2+[2(根(2ab))+2(根(2bc))]^2=4ab+8ac+8bc+16c根(ab)于是,[(a+b)^2+(a+
a、b、c为正实数,求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具体解释啊``]
解:由均值不等式,得
(a+b)^2+(a+b+4c)^2
=(a+b)^2+[(a+2c)+(b+2c)]^2
>=(2根ab)^2+[2(根(2ab))+2(根(2bc))]^2
=4ab+8ac+8bc+16c根(ab)
于是,[(a+b)^2+(a+b+4c)^2](a+b+c)/abc
>=[4ab+8ac+8bc+16c(ab)](a+b+c)/abc
=(4/c+8/b+8/a+16/根ab)(a+b+c)
=8(1/2c+1/b+1/a+1/根ab+1/根ab)(a/2+a/2+b/2+b/2+c)
>=8[5(1/2a^2b^2c)^(1/5)]×[5(a^2b^2c/2^4)^(1/5)]=100
当且仅当a=b=2c>0时,上式取等号,
故原式最小值为100.
这是我找到的那个答案,
但是,
=(4/c+8/b+8/a+16/根ab)(a+b+c)①
=8(1/2c+1/b+1/a+1/根ab+1/根ab)(a/2+a/2+b/2+b/2+c)②
>=8[5(1/2a^2b^2c)^(1/5)]×[5(a^2b^2c/2^4)^(1/5)]=100③
①到②为什么整理成这种形式,②到③是怎么得的,有没有用什么公式,怎么出来一个五次根下?puzzle
a、b、c为正实数,求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具体解释啊``]解:由均值不等式,得(a+b)^2+(a+b+4c)^2=(a+b)^2+[(a+2c)+(b+2c)]^2>=(2根ab)^2+[2(根(2ab))+2(根(2bc))]^2=4ab+8ac+8bc+16c根(ab)于是,[(a+b)^2+(a+
=(4/c+8/b+8/a+16/根ab)(a+b+c)①
=8(1/2c+1/b+1/a+1/根ab+1/根ab)这个括号是提取公因数,没什么好说的.
对于(a/2+a/2+b/2+b/2+c)那是因为第一步等号成立已经限定了a=b=2c,故这个括号里的项也必须分称这种形式,并且项数和第一个括号一致,才能齐次.
公式很简单,就是a+b+c+d≥4(abcd)^(1/4)]这里的4和前面有几个数是对应的,对于n都成立