已经a+b=1[(1/a)+1]*[(1/b)+1]=(a+b/a +a/a)(a+b/a +b/b)=(2a+b)(2b+a)/ab2a+b≥2*根号2ab2b+a≥2*根号2ba(2a+b)(2b+a)≥8ab所以(2a+b)(2b+a)/ab≥8,=(a+1)(b+1)/ab=ab+a+b+1/ab=ab+2/ab=2(a+b)/ab +1a+b≥2*根号ab(a+b)^2/4≥ab1/ab≥4a+b/ab≥4所
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![已经a+b=1[(1/a)+1]*[(1/b)+1]=(a+b/a +a/a)(a+b/a +b/b)=(2a+b)(2b+a)/ab2a+b≥2*根号2ab2b+a≥2*根号2ba(2a+b)(2b+a)≥8ab所以(2a+b)(2b+a)/ab≥8,=(a+1)(b+1)/ab=ab+a+b+1/ab=ab+2/ab=2(a+b)/ab +1a+b≥2*根号ab(a+b)^2/4≥ab1/ab≥4a+b/ab≥4所](/uploads/image/z/11281815-63-5.jpg?t=%E5%B7%B2%E7%BB%8Fa%2Bb%3D1%5B%281%2Fa%29%2B1%5D%2A%5B%281%2Fb%29%2B1%5D%3D%28a%2Bb%2Fa+%2Ba%2Fa%29%28a%2Bb%2Fa+%2Bb%2Fb%29%3D%282a%2Bb%29%282b%2Ba%29%2Fab2a%2Bb%E2%89%A52%2A%E6%A0%B9%E5%8F%B72ab2b%2Ba%E2%89%A52%2A%E6%A0%B9%E5%8F%B72ba%282a%2Bb%29%282b%2Ba%29%E2%89%A58ab%E6%89%80%E4%BB%A5%282a%2Bb%29%282b%2Ba%29%2Fab%E2%89%A58%2C%3D%28a%2B1%29%28b%2B1%29%2Fab%3Dab%2Ba%2Bb%2B1%2Fab%3Dab%2B2%2Fab%3D2%28a%2Bb%29%2Fab+%2B1a%2Bb%E2%89%A52%2A%E6%A0%B9%E5%8F%B7ab%28a%2Bb%29%5E2%2F4%E2%89%A5ab1%2Fab%E2%89%A54a%2Bb%2Fab%E2%89%A54%E6%89%80)
已经a+b=1[(1/a)+1]*[(1/b)+1]=(a+b/a +a/a)(a+b/a +b/b)=(2a+b)(2b+a)/ab2a+b≥2*根号2ab2b+a≥2*根号2ba(2a+b)(2b+a)≥8ab所以(2a+b)(2b+a)/ab≥8,=(a+1)(b+1)/ab=ab+a+b+1/ab=ab+2/ab=2(a+b)/ab +1a+b≥2*根号ab(a+b)^2/4≥ab1/ab≥4a+b/ab≥4所
已经a+b=1
[(1/a)+1]*[(1/b)+1]
=(a+b/a +a/a)(a+b/a +b/b)
=(2a+b)(2b+a)/ab
2a+b≥2*根号2ab
2b+a≥2*根号2ba
(2a+b)(2b+a)≥8ab
所以(2a+b)(2b+a)/ab≥8
,=(a+1)(b+1)/ab
=ab+a+b+1/ab
=ab+2/ab
=2(a+b)/ab +1
a+b≥2*根号ab
(a+b)^2/4≥ab
1/ab≥4
a+b/ab≥4
所以2(a+b)/ab +1≥9
第1个错在哪?
已经a+b=1[(1/a)+1]*[(1/b)+1]=(a+b/a +a/a)(a+b/a +b/b)=(2a+b)(2b+a)/ab2a+b≥2*根号2ab2b+a≥2*根号2ba(2a+b)(2b+a)≥8ab所以(2a+b)(2b+a)/ab≥8,=(a+1)(b+1)/ab=ab+a+b+1/ab=ab+2/ab=2(a+b)/ab +1a+b≥2*根号ab(a+b)^2/4≥ab1/ab≥4a+b/ab≥4所
第1个:
2a+b≥2*根号2ab :2a=b,即:a=1/3,b=2/3时等号成立;
2b+a≥2*根号2ba :a=2b,即:a=2/3,b=1/3时等号成立;
两式并不能同时得到最小值
所以,他们的乘积也不能得到最小值
即:
(2a+b)(2b+a)≥8ab 中的等号是不会成立的
第一个两次均值时取等条件不一致,因此取不到等号
已经a+b=1
[(1/a)+1]*[(1/b)+1]
=(a+b/a +a/a)(a+b/a +b/b)
=(2a+b)(2b+a)/ab
2a+b≥2*根号2ab
2b+a≥2*根号2ba
(2a+b)(2b+a)≥8ab
所以(2a+b)(2b+a)/ab≥8
2a+b≥2*根号2ab
2b+a≥2*根号2ba
不可能同时取等!!!!!!!