a、b、c為正實(shí)數(shù),求[(a+b)^2+(a+b+4c)^2](a+b+c)/abc的最小值. [求高手具體解釋啊``]
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
當(dāng)且僅當(dāng)a=b=2c>0時(shí),上式取等號(hào),
故原式最小值為100.
這是我找到的那個(gè)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為什么整理成這種形式,②到③是怎么得的,有沒有用什么公式,怎么出來一個(gè)五次根下?puzzle