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\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}\)
\(A=\frac{ac+c+1}{ac+c+1}\)
\(A=1\)
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\(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)
\(=\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{1+bc+b}\)
\(=\frac{1}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{1.a}{a.\left(1+bc+b\right)}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{a+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{a}{ab+a+1}=\frac{ab+a+1}{ab+a+1}=1\)
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Ta có: \(0\le a\le b\le1\Rightarrow\hept{\begin{cases}a-1\ge0\\b-1\ge0\end{cases}}\)
\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab-a-b+1\ge0\)
\(\Leftrightarrow ab+1\ge a+b\Leftrightarrow\frac{c}{ab+1}\le\frac{c}{a+b}\)(Vì \(c\ge0\))
Mà \(\frac{c}{a+b}\le\frac{c+c}{a+b+c}=\frac{2c}{a+b+c}\)(Vì \(c\ge0\))
\(\Rightarrow\frac{c}{ab+1}\le\frac{2c}{a+b+c}\)
Chứng minh tương tự: \(\frac{b}{bc+1}\le\frac{2b}{a+b+c};\frac{c}{ab+1}\le\frac{2c}{a+b+c}\)
\(\Rightarrow\frac{a}{bc+1}+\frac{b}{bc+1}+\frac{c}{ab+1}\le\frac{2\left(a+b+c\right)}{a+b+c}=2\left(đpcm\right)\)
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Có : 1/ab+a+1 = abc/ab+a+abc = bc/b+1+bc
1/abc+bc+b = 1/1+bc+b
=> 1/ab+a+1 + b/bc+b+1 + 1/abc+bc+b = bc/bc+c+1 + b/bc+b+1 + 1/bc+b+1 = bc+b+1/bc+b+1 = 1
=> ĐPCM
k mk nha
Có : 1/ab+a+1 = abc/ab+a+abc = bc/b+1+bc
1/abc+bc+b = 1/1+bc+b
=> 1/ab+a+1 + b/bc+b+1 + 1/abc+bc+b = bc/bc+c+1 + b/bc+b+1 + 1/bc+b+1 = bc+b+1/bc+b+1 = 1
=> ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
Yêu cầu của đề là gì vậy ?