Toán a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 + 1 >= a + b + c + d 18/07/2021 By Kaylee a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 + 1 >= a + b + c + d
ta có :`(2a-1)^2+(2b-1)^2+(2c-1)^2+(2d-1)^2≥0` `⇔(4a^2-4a+1)+(4b^2-4b+1)+(4c^2-4c+1)+(4d^2-4d+1)≥0` `⇔4a ^ 2 +4b ^ 2 +4c ^ 2 +4d ^ 2 + 4-4a – 4b – 4c – 4d≥0` `⇔4a ^ 2 +4b ^ 2 +4c ^ 2 +4d ^ 2 + 4≥4a + 4b + 4c + 4d` `⇔a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 + 1≥a + b + c + d` Trả lời
ta có :
`(2a-1)^2+(2b-1)^2+(2c-1)^2+(2d-1)^2≥0`
`⇔(4a^2-4a+1)+(4b^2-4b+1)+(4c^2-4c+1)+(4d^2-4d+1)≥0`
`⇔4a ^ 2 +4b ^ 2 +4c ^ 2 +4d ^ 2 + 4-4a – 4b – 4c – 4d≥0`
`⇔4a ^ 2 +4b ^ 2 +4c ^ 2 +4d ^ 2 + 4≥4a + 4b + 4c + 4d`
`⇔a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 + 1≥a + b + c + d`