Cho `a_1 , a_2 , …. , a_n > 0` . CMR :
`a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 >= a_1/a_2 + a_2/a_3 + … + a_n/a_1`
Cho `a_1 , a_2 , …. , a_n > 0` . CMR :
`a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 >= a_1/a_2 + a_2/a_3 + … + a_n/a_1`
Đáp án:
Ta có
`a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 >= a_1/a_2 + a_2/a_3 + … + a_n/a_1`
`<=> a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2- a_1/a_2 – a_2/a_3 – … – a_n/a_1 >= 0`
`<=> a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 – 2a_1/a_2 – 2a_2/a_3 – … – 2a_n/a_1 + a_1/a_2 + a_2/a_3 + … + a_n/a_1 >= 0 (1)`
`VT (1) >= a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 – 2a_1/a_2 – 2a_2/a_3 – … – 2a_n/a_1 + n = (a_1^2/a_2^2 – 2a_1/a_2 + 1) + …. + (a_n^2/a_1^2 – 2a_n/a_1 + 1) = (a_1/a_2 – 1)^2 + … + (a_n/a_1 – 1)^2 >= 0 = VP (1) (đpcm)`
Dấu “=” xảy ra `<=> a_1/a_2 = a_2/a_3 = …. = a_n/a_1 = 1 <=> a_1 = a_2 = …. = a_n`
hoặc bn có thể làm như sau:
`(a_1/a_2 – 1)^2 >= 0 <=> a_1^2/a_2^2 >= 2a_1/a_2 – 1`
`(a_2/a_3 – 1)^2 >= 0 <=> a_2^2/a_3^2 >= 2a_2/a_3 – 1`
`………….`
`(a_n/a_1 – 1)^2 >= 0 <=> a_n^2/a_1^2 >= 2a_n/a_1 – 1`
Cộng từng vế lại ta được
`a_1^2/a_2^2 + a_2^2/a_3^2 + …. + a_n^2/a_1^2 >= 2(a_1/a_2 + a_2/a_3 + … + a_n/a_1) – n = (a_1/a_2 + a_2/a_3 + … + a_n/a_1) + (a_1/a_2 + a_2/a_3 + … + a_n/a_1) – n >= a_1/a_2 + a_2/a_3 + … + a_n/a_1 + n – n = a_1/a_2 + a_2/a_3 + … + a_n/a_1 (đpcm)`
Dấu “=” xảy ra `<=> a_1 = a_2 = … = a_n`
Giải thích các bước giải: