cho `a;b;c`lớn hơn ko `a+b+c=4` `cmr:A=(a)/(b^6+729)+(b)/(c^6+729)+(c)/(a^6+729)` 11/07/2021 Bởi Sarah cho `a;b;c`lớn hơn ko `a+b+c=4` `cmr:A=(a)/(b^6+729)+(b)/(c^6+729)+(c)/(a^6+729)`
mình làm hơi tắt `a/(b^6+729)=1/(729)(a-a^6/(b^6+729))≥1/(729)(a-(ab^6)/(54b^3))=1/(729)(a-(ab^3)/(54))` tương tự `:` `b/(c^6+729)≥1/(729)(b-(bc^3)/(54))` `c/(a^6+729)≥1/(729)(c-(ca^3)/(54))` `⇒A≥1/(729)(a-(ab^3)/(54)+b-(bc^3)/(54)+c-(ca^3)/(54))` `⇒A≥1/(729)(4-(ab^3+bc^3+ca^3)/(54))` (lats mình cm ở https://hoidap247.com/cau-hoi/2031142) đây nha bạn `⇒A≥1/(729)(4-(27)/(54))` `⇒A≥(7)/(1458)` `”=”`xẩy ra khi : `(a;b;c)=(0;1;3);(1;3;0);(3;0;1)` Bình luận
mình làm hơi tắt
`a/(b^6+729)=1/(729)(a-a^6/(b^6+729))≥1/(729)(a-(ab^6)/(54b^3))=1/(729)(a-(ab^3)/(54))`
tương tự `:`
`b/(c^6+729)≥1/(729)(b-(bc^3)/(54))`
`c/(a^6+729)≥1/(729)(c-(ca^3)/(54))`
`⇒A≥1/(729)(a-(ab^3)/(54)+b-(bc^3)/(54)+c-(ca^3)/(54))`
`⇒A≥1/(729)(4-(ab^3+bc^3+ca^3)/(54))` (lats mình cm ở https://hoidap247.com/cau-hoi/2031142) đây nha bạn
`⇒A≥1/(729)(4-(27)/(54))`
`⇒A≥(7)/(1458)`
`”=”`xẩy ra khi :
`(a;b;c)=(0;1;3);(1;3;0);(3;0;1)`