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\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{1}{b+1+bc}+\dfrac{1}{c+1+ac}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac}{abc+ac+abc.c}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac}{1+ac+c}+\dfrac{1}{ac+c+c}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac+1+c}{ac+c+1}=1\) (đpcm)

Lời giải:

Vì \(a,b,c\in [0;1]\Rightarrow b^2\leq b; c^3\leq c\)

\(\Rightarrow a+b^2+c^3-ab-bc-ac\leq a+b+c-ab-bc-ac(*)\)

\(a,b,c\leq 1\Rightarrow (a-1)(b-1)(c-1)\le 0\)

\(\Leftrightarrow (ab-a-b+1)(c-1)\leq 0\)

\(\Leftrightarrow abc-(ab+bc+ac)+(a+b+c)-1\leq 0\)

\(\Leftrightarrow a+b+c-(ab+bc+ac)\leq 1-abc\leq 1(**)\) (do $abc\geq 0$)

Từ \((*); (**)\Rightarrow a+b^2+c^3-ab-bc-ac\leq 1\)

Ta có đpcm.

Lời giải:
Dựa vào điều kiện $abc=1$ ta có:

\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+ca+c}=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{1+ca+c}\)

\(=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{ab+ab.ca+ab.c}\)

\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{ab+a+1}=\frac{1+a+ab}{ab+a+1}=1\)

Ta có đpcm.

Ta có: \(a.b.c=1\)

\(=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+bc+b}\)

\(=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{abc.a+abc+ab}\)

\(=\frac{1}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{a}{a+1+ab}\)

\(=\frac{1+ab+a}{1+ab+a}\)

\(=1.\)

\(\Rightarrow\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+bc+b}=1\left(đpcm\right).\)

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