cho a,b,c>0 a.b.c=1 Tìm MAX M=1/(ab+a+2)+1/(bc+b+2)+1/(ca+c+2)

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1. `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`

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2. 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

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