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

\(A=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

\(A=\dfrac{a^2bc}{ab\left(1+ac+c\right)}+\dfrac{b}{b\left(c+1+ac\right)}+\dfrac{c}{ac+c+1}\)

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

\(A=1\)

\(A=\dfrac{ab}{ab+a+1}+\dfrac{bc}{bc+b+1}+\dfrac{ca}{ca+c+1}\)

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

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

\(A=1\)

Lười đánh máy thật sự, buốt tay lắm:((

Ta có: \(Q=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

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

\(Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}\)

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

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

Vậy Q=1

Q=ab+a+1a​+bc+b+1b​+ac+c+1c​

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

Q=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abc^2+abc+ac}+\dfrac{c}{ac+c+1}Q=abc+ac+cac​+abc2+abc+acabc​+ac+c+1c​

Q=\dfrac{ac}{1+ac+c}+\dfrac{1}{c+a+ac}+\dfrac{c}{ac+c+1}Q=1+ac+cac​+c+a+ac1​+ac+c+1c​

Q=\dfrac{ac+1+c}{1+ac+c}=1Q=1+ac+cac+1+c​=1

chúc bạn thi tốt

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

\(\Leftrightarrow M=\dfrac{1}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{a}{a+1+ab}\)

\(\Leftrightarrow M=\dfrac{1+ab+a}{ab+a+1}=1\)

\(VT=\dfrac{a^3bc}{c+ab^2c}+\dfrac{ab^3c}{a+abc^2}+\dfrac{abc^3}{b+a^2bc}\)

\(=abc\left(\dfrac{a^2}{c+ab^2c}+\dfrac{b^2}{a+abc^2}+\dfrac{c^2}{b+a^2bc}\right)\)

Áp dụng bđt Cauchy-Schwarz dạng engel có:

\(VT\ge\dfrac{abc\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}\)\(=\dfrac{abc\left(a+b+c\right)}{1+abc}\)

Dấu "=" xảy ra khi \(a=b=c\)

Vậy...

Sai đề không bạn,tại a=b=c=2 thay vào không thỏa mãn nha

\(\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{abc+ab+a}+\dfrac{2c}{ac+2c+abc}\)

\(=\dfrac{a}{ab+a+2}+\dfrac{ab}{2+ab+a}+\dfrac{2}{a+2+ab}\)

\(=\dfrac{ab+a+2}{ab+a+2}=1\)

Lời giải:

Vì $a+b+c=1$ nên:

\(a^2+b^2+abc-1=(a+b)^2-2ab+abc-1\)

\(=(a+b)^2-1+ab(c-2)=(1-c)^2-1+ab(c-2)\)

\(=-c(2-c)+ab(c-2)=c(c-2)+ab(c-2)=(c+ab)(c-2)\)

Do đó:

\(\frac{c+ab}{a^2+b^2+abc-1}=\frac{c+ab}{(c+ab)(c-2)}=\frac{1}{c-2}\)

Hoàn toàn tương tự với các phân thức còn lại, suy ra:

\(\frac{c+ab}{a^2+b^2+abc-1}+\frac{a+bc}{b^2+c^2+abc-1}+\frac{b+ac}{a^2+c^2+abc-1}=\frac{1}{c-2}+\frac{1}{a-2}+\frac{1}{b-2}=\frac{(a-2)(b-2)+(b-2)(c-2)+(c-2)(a-2)}{(a-2)(b-2)(c-2)}\)

\(=\frac{ab+bc+ac-4(a+b+c)+12}{(a-2)(b-2)(c-2)}=\frac{ab+bc+ac+8}{(a-2)(b-2)(c-2)}\)

Ta có đpcm.

Akai Haruma

Do \(abc=1\Rightarrow\) đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

\(VT=\dfrac{xz}{y\left(x+z\right)}+\dfrac{xy}{z\left(x+y\right)}+\dfrac{yz}{x\left(y+z\right)}=\dfrac{\left(xz\right)^2}{xyz\left(x+z\right)}+\dfrac{\left(xy\right)^2}{xyz\left(x+y\right)}+\dfrac{\left(yz\right)^2}{xyz\left(y+z\right)}\)

\(VT\ge\dfrac{\left(xy+yz+zx\right)^2}{2xyz\left(x+y+z\right)}\ge\dfrac{3xyz\left(x+y+z\right)}{2xyz\left(x+y+z\right)}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(x=y=z\) hay \(a=b=c=1\)

Ta có :

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

\(A=\dfrac{a}{ab+a+1}+\dfrac{ab}{abc+ab+a}+\dfrac{abc}{aabc+abc+ab}\)

\(A=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{1}{a+1+ab}\)

\(A=\dfrac{a+ab+1}{ab+a+1}\)

\(\Rightarrow A=1\left(đpcm\right)\)

kiểm tra lại đề đi bạn