Ta có :

\(A+3=\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+3\)

\(=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)

\(=2017.\frac{1}{2017}=1\)

\(\Rightarrow A=1-3=-2\)

Đặt : \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=P\)

\(\Rightarrow\left(a+b+c\right).P=\frac{1}{2019}.2019\)

\(\Rightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{c+a}=\frac{6057}{2019}+\frac{\left(-4038\right)}{2019}\)

\(\Rightarrow3+\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=3+\left(-2\right)\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=-2\)

cảm ơn bạn nhé

Maknae W1 _ chungha _ ioi làm hơi dài dòng

\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{a+c}+1\right)+\left(\frac{c}{a+b}+1\right)\)

\(S+3=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}\)

\(S+3=\left(a+b+c\right).\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{2019.1}{2019}=1\Rightarrow S=1-3=-2\)

cách này đỡ phải tính số lớn :")

ta có \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+c}\)

=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{b+c}+1-3\)

=\(\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)

=\(\left(a+b+c\right)\left(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)

rồi còn lại thay vào nha bn

\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)

\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=1\)

\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+3=1\)

\(\Leftrightarrow S=-2\)

\(S=\left(\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\right)-3\)

\(S=\left(a+b+c\right).\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(S=\frac{2019.1}{2019}-3\)

\(S=1-3=-2\)

a) a = 2 , b = 3, c = 6

\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)

\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)

\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)

Mà \(a^2+b^2+c^2\ge ab+bc+ca\Rightarrowđpcm\)

\(\frac{a^3}{b+c}+\frac{a^3}{b+c}+\frac{\left(b+c\right)^2}{8}\ge3\sqrt[3]{\frac{a^3}{b+c}.\frac{a^3}{b+c}.\frac{\left(b+c\right)^2}{8}}=\frac{3a^2}{2}\)

Rồi tương tự các kiểu:v

Suy ra \(2VT\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{8}\)

\(\ge\frac{3}{2}\left(a^2+b^2+c^2\right)-\frac{a^2+b^2+c^2}{2}=\left(a^2+b^2+c^2\right)\) (chú ý \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))

Không phải dùng tới Cauchy-Schwarz:D

mình chưa hiểu?

có thể giải thích rõ hơn đc ko

Ok , mình sẽ làm !

Ta có :

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b}{c}-1+1=\frac{b+c}{a}-1+1=\frac{c+a}{b}-1+1\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\left(1\right)\)

+) Trường hợp 1 : \(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

Ta có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-a}{a}.\frac{-c}{c}.\frac{-b}{b}\)

\(\Leftrightarrow P=-1.\left(-1\right).\left(-1\right)=-1\)

+) Trường hợp 2 : \(a+b+c\ne0\)

Áp dụng tính chất của dãy tỉ số bằng nhau cho ( 1 ) , ta có :

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

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Ta lại có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(\Leftrightarrow P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{c+b}{b}\)

\(\Leftrightarrow P=2.2.2=8\)

Vậy....................

Đề sai nhé bạn ! Bạn kiểm tra lại!

Sửa lại rồi đó bạn ơi

=> (a+b+c).(1/a+b + 1/b+c  +1/c+a) = 2017/90

=> a+b+c/a+b + a+b+c/b+c + a+b+c/c+a = 2017/90

=> 1 + c/a+b + 1 + a/b+c + 1 + b/c+a = 2017/90

=> a/b+c + b/c+a  +c/a+b = 2017/90 - 3 = 1747/90

Vậy S = 1747/90

Tk mk nha

Ta có: \(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(3+S=\left(1+\frac{a}{b+c}\right)+\left(1+\frac{b}{c+a}\right)+\left(1+\frac{c}{a+b}\right)\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(=2007.\frac{1}{90}=\frac{223}{10}\Rightarrow S=\frac{223}{10}-3=\frac{193}{10}\)

\(S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

\(=>S+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)

\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)

\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{c}{a+b}\right)\)

\(=2007.\frac{1}{90}=\frac{223}{10}\)

\(=>S=\frac{223}{10}-\frac{30}{10}=\frac{193}{10}\)