Ap dông B§T C-S ta cã:

\(\frac{a}{a+\sqrt{2016a+bc}}=\frac{a}{a+\sqrt{\left(a+b+c\right)a+bc}}=\frac{a}{a+\sqrt{\left(a+b\right)\left(c+a\right)}}\)

\(\le\frac{a}{a+\sqrt{\left(\sqrt{ab}+\sqrt{ac}\right)^2}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}\)

\(=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\). T­uong tù ta cx cã: 

\(\frac{b}{b+\sqrt{2016b+ca}}\le\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}};\frac{c}{c+\sqrt{2016c+ab}}\le\frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)

Céng theo vÕ c¸c B§T trªn ta dc:

\(VT\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)

P/s:may mk bi loi Unikey r` mk dg ban chua kip chinh lai bn gang doc 

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

`(2bc-2016)/(3c-2bc+2016)`

`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`

`=-1+(3c)/(3c-2bc+2016)`

`(2b)/(3-2b+ab)`

`=(2bc)/(3c-2bc+abc)`

`=(2bc)/(3c-2bc+2016)`

`(4032-3ac)/(3ac-4032+2016a)`

`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`

`=-1+(2016a)/(3ac-2abc+2016a)`

`=-1+(2016)/(3c-2bc+2016)`

`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)`

`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`

`=>M=-2+1`

`=>M=-1`

Nãy thiếu latex ạ sorry~~

giúp với m.n

Làm đơn giản thế này thôi nhé An Kì :

Ta có : \(2016a+bc=\left(a+b+c\right)a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)Tương tự : \(2016b+ac=\left(a+b\right)\left(b+c\right)\)

\(2016c+ab=\left(a+c\right)\left(b+c\right)\)

\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a}{a+\sqrt{2016a + bc}}=\frac{a}{a+\sqrt{(a+b+c)a + bc}} =\frac{a}{a+\sqrt{(a+b)(c+a)}} \leq \frac{a}{a+\sqrt{(\sqrt{ab}+\sqrt{ac})^{2}}}=\frac{a}{a+\sqrt{ab}+\sqrt{ac}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$

$\Rightarrow \frac{a}{a+\sqrt{2016a + bc}} + \frac{b}{b+\sqrt{2016b + ca}} + \frac{c}{c+\sqrt{2016c + ab}}\leq \frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$

...............................

Tớ ko bt

\(A=\frac{2016a}{ab+2016a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}\)

\(A=\frac{2016a}{ab+2016a+abc}+\frac{b}{bc+b+2016}+\frac{bc}{abc+bc+b}\)

\(A=\frac{2016a}{a\left(b+2016+bc\right)}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016}{b+2016+bc}+\frac{b}{bc+b+2016}+\frac{bc}{2016+bc+b}\)

\(A=\frac{2016+b+bc}{2016+b+bc}=1\)

Thay : 2016 = abc

ta có :

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

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

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

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

\(A=1\)

vậy \(A=\frac{2016.a}{ab+2016.a+2016}+\frac{b}{bc+b+2016}+\frac{c}{ac+c+1}=1\)

Chúc bạn học tốt !

Công dãy lại => hệ số : \(k=2014\)

Cách đơn giảii không hiệu quả, Thế lại=> a,b,c thay vào ra A