Lời giải:

Vế đầu:

Áp dụng BĐT AM-GM:

$(ab+bc+ac)(a+b+c)\geq 9abc$

$\Leftrightarrow ab+bc+ac\geq 9abc$

$\Rightarrow ab+bc+ac-2abc\geq 9abc-2abc=7abc\geq 0$ do $a,b,c\geq 0$

Vế sau:

Áp dụng BĐT Schur:

$abc\geq (a+b-c)(b+c-a)(c+a-b)=(1-2a)(1-2b)(1-2c)$

$\Leftrightarrow 9abc\geq 4(ab+bc+ac)-1$

$\Rightarrow 2abc\geq \frac{8}{9}(ab+bc+ac)-\frac{2}{9}$

$\Rightarrow ab+bc+ac-2abc\leq ab+bc+ac-[\frac{8}{9}(ab+bc+ac)-\frac{2}{9}]=\frac{ab+bc+ac}{9}+\frac{2}{9}$

$\leq \frac{(a+b+c)^2}{27}+\frac{2}{9}$ (theo BĐT AM-GM)

$=\frac{1}{27}+\frac{2}{9}=\frac{7}{27}$

Ta có đpcm.

BĐT bị ngược dấu, BĐT đúng phải là:

\(\dfrac{a}{ac+4}+\dfrac{b}{ab+4}+\dfrac{c}{bc+4}\le\dfrac{a^2+b^2+c^2}{16}\)

k mk đi

ai k mk 

mk k lại

thanks

Vì \(0\le a\le b\le c\le1\) nên:

\(\left(a-1\right)\left(b-1\right)\ge ab+1\ge a+b\Leftrightarrow\dfrac{1}{ab+1}\le\dfrac{1}{a+b}\Leftrightarrow\dfrac{c}{ab+1}\le\dfrac{c}{a+b}\left(1\right)\)

Tương tự: \(\dfrac{a}{bc+1}\le\dfrac{a}{b=c}\left(2\right);\dfrac{b}{ac+1}\le\dfrac{b}{a+c}\left(3\right)\)

Do đó: \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\le\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\left(4\right)\)

Mà: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\le\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(5\right)\)

Từ (4) và (5) suy ra \(\dfrac{a}{bc+1}+\dfrac{b}{ac+1}+\dfrac{c}{ab+1}\left(đpcm\right)\)

a.

\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)

2.

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)

Quay lại câu a

\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)

\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)

Dấu $"="$ khi $a=b=c$

Áp dụng bđt cosi ta có:

`a+b>=2sqrt{ab}`

`=>(ab)/(a+b)<=(sqrt{ab})/2`

Chứng minh tt:

`(bc)/(b+c)<=(sqrt{bc})/2`

`(ca)/(a+c)<=(sqrt{ca})/2`

`=>VT<=(sqrt{ab}+sqrt{bc}+sqrt{ca})/2`

Áp dụng cosi:

`sqrt{ab}<=(a+b)/2`

`sqrt{bc}<=(b+c)/2`

`sqrt{ca}<=(c+a)/2`

`=>(sqrt{ab}+sqrt{bc}+sqrt{ca})/2<=(a+b+c)/2`

`=>VT<=(a+b+c)/2`