Toán cho a,b,c>0 a.b.c=1 Tìm MAX M=1/(ab+a+2)+1/(bc+b+2)+1/(ca+c+2) 15/07/2021 By Aaliyah cho a,b,c>0 a.b.c=1 Tìm MAX M=1/(ab+a+2)+1/(bc+b+2)+1/(ca+c+2)
`1/(ab+a+2)≤1/4 (1/(ab+1)+1/(a+1))` tương tự `1/(cb+b+2)≤1/4 (1/(cb+1)+1/(b+1))` `1/(ac+c+2)≤1/4 (1/(ac+1)+1/(c+1))` `⇒M≤1/4 (1/(ac+1)+1/(c+1))+1/4 (1/(cb+1)+1/(b+1))+1/4 (1/(ab+1)+1/(a+1))` `⇒M≤1/4 ((ab+1)/(ab+1)+(ac+1)/(ac+1)+(bc+1)/(bc+1))` `⇒M≤1/4 .3≤3/4` `”=”`xẩy ra khi : `a=b=c=1` Trả lời
M=$\frac{1}{ab+a+2}$ +$\frac{1}{bc+b+2}$ +$\frac{1}{ca+c+2}$ Ta có: $\frac{1}{ab+a+2}$ =$\frac{1}{(ab+1)+(a+1)}$ ≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$) (BĐT Bunhicopski) Tương tự: $\frac{1}{bc+b+2}$ ≤$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$) $\frac{1}{ca+c+2}$ ≤$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$) ⇒M≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$)+$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$)+$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$+$\frac{1}{bc+1}$+$\frac{1}{b+1}$+$\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{ab+1}{ab+1}$ +$\frac{bc+1}{bc+1}$ +$\frac{ca+1}{ca+1}$) =$\frac{1}{4}$(1+1+1)=$\frac{3}{4}$ Dấu “=” xảy ra a=b=c=1 Vậy Max M=$\frac{3}{4}$ khi a=b=c=1 Trả lời
`1/(ab+a+2)≤1/4 (1/(ab+1)+1/(a+1))`
tương tự
`1/(cb+b+2)≤1/4 (1/(cb+1)+1/(b+1))`
`1/(ac+c+2)≤1/4 (1/(ac+1)+1/(c+1))`
`⇒M≤1/4 (1/(ac+1)+1/(c+1))+1/4 (1/(cb+1)+1/(b+1))+1/4 (1/(ab+1)+1/(a+1))`
`⇒M≤1/4 ((ab+1)/(ab+1)+(ac+1)/(ac+1)+(bc+1)/(bc+1))`
`⇒M≤1/4 .3≤3/4`
`”=”`xẩy ra khi :
`a=b=c=1`
M=$\frac{1}{ab+a+2}$ +$\frac{1}{bc+b+2}$ +$\frac{1}{ca+c+2}$
Ta có:
$\frac{1}{ab+a+2}$ =$\frac{1}{(ab+1)+(a+1)}$ ≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$) (BĐT Bunhicopski)
Tương tự:
$\frac{1}{bc+b+2}$ ≤$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$)
$\frac{1}{ca+c+2}$ ≤$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$)
⇒M≤$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$)+$\frac{1}{4}$($\frac{1}{bc+1}$+$\frac{1}{b+1}$)+$\frac{1}{4}$($\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{1}{ab+1}$+$\frac{1}{a+1}$+$\frac{1}{bc+1}$+$\frac{1}{b+1}$+$\frac{1}{ca+1}$+$\frac{1}{c+1}$)=$\frac{1}{4}$($\frac{ab+1}{ab+1}$ +$\frac{bc+1}{bc+1}$ +$\frac{ca+1}{ca+1}$) =$\frac{1}{4}$(1+1+1)=$\frac{3}{4}$
Dấu “=” xảy ra a=b=c=1
Vậy Max M=$\frac{3}{4}$ khi a=b=c=1