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Vì: \(0\le a\le b\le c\le1\) nên:
\(\left(a-1\right).\left(b-1\right)\ge0\Leftrightarrow ab-a-b+1\ge0\Leftrightarrow ab+1\ge a+b\)
\(\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\) (1)
\(\left(a-1\right).\left(c-1\right)\ge0\Leftrightarrow ac-a-c+1\ge0\Leftrightarrow ac+1\ge a+c\)
\(\Leftrightarrow\dfrac{1}{ac+1}\le\dfrac{1}{a+c}\Leftrightarrow\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\) (2)
\(\left(b-1\right).\left(c-1\right)\ge0\Leftrightarrow bc-b-c+1\ge0\Leftrightarrow bc+1\ge b+c\)
\(\Leftrightarrow\dfrac{1}{bc+1}\le\dfrac{1}{b+c}\Leftrightarrow\dfrac{a}{bc+1}\le\dfrac{a}{b+c}\) (3)
Cộng vế với vế của (1)(2) và (3) ta được:
\(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2a+2b+2c}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{2.\left(a+b+c\right)}{a+b+c}\)
\(\Leftrightarrow\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ac+1}\le2\left(đpcm\right)\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)
\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{ab.ac+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1\)
cho ba số a,b,c thỏa mãn a.b.c = 1 . CMR: \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+bc+b}=1\)
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: \(a.b.c=1\)
\(=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+bc+b}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{abc.a+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{a}{a+1+ab}\)
\(=\frac{1+ab+a}{1+ab+a}\)
\(=1.\)
\(\Rightarrow\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{abc+bc+b}=1\left(đpcm\right).\)
Chúc bạn học tốt!
\(A=\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)
\(A=\frac{c}{ac+c+1}+\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}\)
\(A=\frac{ac+c+1}{ac+c+1}\)
\(A=1\)