Chắc bạn viết nhầm đề, cho \(a=b=c=1\) đâu có đúng

Sửa lại đề: cho \(abc=1\) chứng minh \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\)

Ta có

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

\(=\dfrac{a}{ab+a+1}+\dfrac{ab}{1+ab+a}+\dfrac{c}{c\left(a+1+ab\right)}\)

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

\(=\dfrac{a+ab+1}{ab+a+1}=1\) (đpcm)

Đề bạn Lâm đúng đấy!

ac+c+1

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

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

\(=\dfrac{ac}{ac+2}+\dfrac{ab}{ab+2}+\dfrac{bc}{bc+2}\)

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

\(=1.1=1\)(đpcm).

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

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

\(=\dfrac{1}{1+a+ab}+\dfrac{a}{1+a+ab}+\dfrac{1}{c\left(1+a+ab\right)}\)

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

Lời giải:

Ta có:

\(\text{VT}=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)

\(=\frac{a.c}{abc+ac+c}+\frac{b.ac}{bc.ac+b.ac+ac}+\frac{c}{ac+c+1}\)

\(=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}\) (thay \(abc=1\) )

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

Ta có đpcm.

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

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

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

\(c\left(1+ab\right)\le c\left(1+\dfrac{a^2+b^2}{2}\right)=c\left(1+\dfrac{1-c^2}{2}\right)=1-\dfrac{1}{2}\left(c-1\right)^2\left(c+2\right)\le1\)

\(\Rightarrow c^2\left(1+ab\right)\le c\Rightarrow\dfrac{c}{1+ab}\ge c^2\)

Hoàn toàn tương tự ta có: \(\dfrac{a}{1+bc}\ge a^2\) ; \(\dfrac{b}{1+ac}\ge b^2\)

Cộng vế: \(VT\ge a^2+b^2+c^2=1\) (đpcm)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị

Cách 2:

Áp dụng BĐT Bunhiacopxky:

\(\text{VT}[a(1+bc)+b(1+ac)+c(1+ab)]\geq (a+b+c)^2\)

\(\Rightarrow \text{VT}\geq \frac{(a+b+c)^2}{a+b+c+3abc}\)

 Ta sẽ CM: 

\(\frac{(a+b+c)^2}{a+b+c+3abc}\geq 1\)

\(\Leftrightarrow 1+2(ab+bc+ac)\geq a+b+c+3abc\)

Vì $a^2+b^2+c^2=1\Rightarrow a,b,c\leq 1$

$\Rightarrow (a-1)(b-1)(c-1)\leq 0$

$\Leftrightarrow 1+ ab+bc+ac\geq a+b+c+abc(1)$

Áp dụng BĐT AM-GM:

$ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}\geq 3\sqrt[3]{a^2b^2c^2.abc}=3abc\geq 2abc(2)$

Từ $(1);(2)\Rightarrow 1+2(ab+bc+ac)\geq a+b+c+3abc$

Ta có đpcm

Dấu "=" xảy ra khi $(a,b,c)=(1,0,0)$ và hoán vị.

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 \(a=b=c=1\)