Cho x + y = 1. Tìm GTNN của B = (1 – 1/x^2)(1 – 1/y^2) 09/11/2021 Bởi Arya Cho x + y = 1. Tìm GTNN của B = (1 – 1/x^2)(1 – 1/y^2)
B = `(1 + 1/x)(1 + 1/y) . (1 – 1/x)(1 – 1/y)``= (1 + 1/x)(1 + 1/y) . (x -1)(y – 1)/(xy)``= (1 + 1/x)(1 + 1/y) . (-x).(-y)/(xy)``= (1 + 1/x)(1 + 1/y)``= 1 + 1/(xy) + (1/x + 1/y) = 1 + 1/(xy) + (x + y)/xy``= 1 + 1/(xy) + 1/(xy) = 1 + 2/(xy)` Áp dụng AM-GM ta có `x+y≥2√xy` `⇒1≥2√xy` `⇒1/2≥√xy` `⇒xy≤1/4` `⇒2/xy≥8` `⇒B≥9` Dấu bằng khi:`x=y=1/2` Bình luận
`B=(1-1/x^2)(1-1/y^2)` `B=(1+1/x)(1+1/y). (1-1/x)(1-1/y)` `B=[(x+1)(y+1)]/(xy). [(x-1)(y-1)]/(xy)` `B=(xy+x+y+1)/(xy). (xy-x-y+1)/(xy)` `B=(xy+2)/(xy). (xy)/(xy)` `B=1+2/(xy)` Theo BĐT $Cô-si$ `xy\le (x+y)^2/4=1/4` `⇒B=1+2/(xy)\le 1+2/(1/4)=9` Dấu `=` xảy ra $\begin{cases}x+y=1\\x=y\end{cases}⇔x=y=\dfrac{1}{2}$ Vậy $Min_B=9⇔x=y=\dfrac{1}{2}$ Bình luận
B = `(1 + 1/x)(1 + 1/y) . (1 – 1/x)(1 – 1/y)`
`= (1 + 1/x)(1 + 1/y) . (x -1)(y – 1)/(xy)`
`= (1 + 1/x)(1 + 1/y) . (-x).(-y)/(xy)`
`= (1 + 1/x)(1 + 1/y)`
`= 1 + 1/(xy) + (1/x + 1/y) = 1 + 1/(xy) + (x + y)/xy`
`= 1 + 1/(xy) + 1/(xy) = 1 + 2/(xy)`
Áp dụng AM-GM ta có
`x+y≥2√xy`
`⇒1≥2√xy`
`⇒1/2≥√xy`
`⇒xy≤1/4`
`⇒2/xy≥8`
`⇒B≥9`
Dấu bằng khi:`x=y=1/2`
`B=(1-1/x^2)(1-1/y^2)`
`B=(1+1/x)(1+1/y). (1-1/x)(1-1/y)`
`B=[(x+1)(y+1)]/(xy). [(x-1)(y-1)]/(xy)`
`B=(xy+x+y+1)/(xy). (xy-x-y+1)/(xy)`
`B=(xy+2)/(xy). (xy)/(xy)`
`B=1+2/(xy)`
Theo BĐT $Cô-si$
`xy\le (x+y)^2/4=1/4`
`⇒B=1+2/(xy)\le 1+2/(1/4)=9`
Dấu `=` xảy ra $\begin{cases}x+y=1\\x=y\end{cases}⇔x=y=\dfrac{1}{2}$
Vậy $Min_B=9⇔x=y=\dfrac{1}{2}$