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

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

\(\Rightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\)

\(\Rightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc=âbc\)

\(\Rightarrow\left(a^2b+ab^2\right)+\left(ac^2+bc^2\right)+\left(a^2c+2abc+b^2c\right)=0\)

\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)

\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+\left(ac+bc\right)\left(a+b\right)=0\)

\(\Rightarrow\left(a+b\right)\left(ab+c^2+ac+bc\right)=0\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=-b\\\frac{b=-c}{a=-c}\end{cases}}\)

Từ đó: P = 0.

Mình giải hơi tắt. Mong bạn hiểu bài.

Chúc bạn học tốt.

Tham khảo ở đây : /hoi-dap/question/77428.html

a/ BĐT sai, cho \(a=b=c=2\) là thấy

b/ \(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)

\(VT\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2}{3\left(a+b+c\right)^2}=\frac{1}{3}\left(a^2+b^2+c^2\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương

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

\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)

Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)

\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)

tc \(0\le a;b;c\le1\)

\(a^3+b^3+c^3+a+b+c=2a^2+2b^2+2c^2=2\)

\(a^3-2a^2+a+b^3-2b^2+b+c^3-2c^2+c=0\)

\(a\left(a-1\right)^2+b\left(b-1\right)^2+c\left(c-1\right)^2=0\)

\(\hept{\begin{cases}a\left(a-1\right)^2=0\\b\left(b-1\right)^2=0\\c\left(c-1\right)^2=0\end{cases}}\)

đến đây lập luận ok

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

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

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

\(=2010.\frac{1}{3}=670\)

\(\Rightarrow S=670-3=667\)

Thêm đk \(a,b,c\ne0\)

Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)

\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)

\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)

\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)

\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)

\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)

a) A:B:C= 4:7:7

=>  A/4 = B/7 = C/7 

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{A}{4}=\frac{B}{7}=\frac{C}{7}=\frac{A+B+C}{4+7+7}=\frac{180o}{18}=10o\)

\(\Rightarrow\frac{A}{4}=10o\Rightarrow A=40o\)

\(\Rightarrow\frac{B}{7}=10o\Rightarrow B=70o\)

\(\Rightarrow\frac{C}{7}=10o\Rightarrow C=70o\)

b) B = 1/2*C => C = 2B

Ta có:  A + B + C = 180o

3/5*B + B + 2B = 180o

(3/5 +1 +2)B =180o

18/5 * B = 180o

B = 180o : 18/5

B= 50o

=> A= 3/5 * B = 3/5 * 50o = 30o

=> C^ = 2B = 2* 50o = 100o

Vậy A =

B=

C=