cho 1/a+1/b+1/c=1/a+b+c tinh P= (a+b)(b^3+c^3)(c^5+a^5)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Rightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)
\(\Rightarrow\left(ab+ac+bc\right)\left(a+b+c\right)=abc\)
\(\Rightarrow a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+3abc=âbc\)
\(\Rightarrow\left(a^2b+ab^2\right)+\left(ac^2+bc^2\right)+\left(a^2c+2abc+b^2c\right)=0\)
\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)
\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+\left(ac+bc\right)\left(a+b\right)=0\)
\(\Rightarrow\left(a+b\right)\left(ab+c^2+ac+bc\right)=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=-b\\\frac{b=-c}{a=-c}\end{cases}}\)
Từ đó: P = 0.
Mình giải hơi tắt. Mong bạn hiểu bài.
Chúc bạn học tốt.
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)
\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S+3=\left(\frac{a}{b+c}+1\right)+\left(\frac{b}{c+a}+1\right)+\left(\frac{c}{a+b}+1\right)\)
\(=\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\)
\(=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(=2010.\frac{1}{3}=670\)
\(\Rightarrow S=670-3=667\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thêm đk \(a,b,c\ne0\)
Ta có: \(\frac{ab}{a+b}=\frac{1}{3}\Rightarrow\frac{a+b}{ab}=3\)
\(\frac{bc}{b+c}=\frac{1}{4}\Rightarrow\frac{bc}{b+c}=4\)
\(\frac{ca}{c+a}=\frac{1}{5}\Rightarrow\frac{c+a}{ca}=5\)
\(\Rightarrow\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{c+a}{ca}=12\)
\(\Leftrightarrow\frac{1}{b}+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{c}=12\)
\(\Leftrightarrow2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=12\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A:B:C= 4:7:7
=> A/4 = B/7 = C/7
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{A}{4}=\frac{B}{7}=\frac{C}{7}=\frac{A+B+C}{4+7+7}=\frac{180o}{18}=10o\)
\(\Rightarrow\frac{A}{4}=10o\Rightarrow A=40o\)
\(\Rightarrow\frac{B}{7}=10o\Rightarrow B=70o\)
\(\Rightarrow\frac{C}{7}=10o\Rightarrow C=70o\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+ac^2-abc=0\)
\(\Leftrightarrow ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)=0\)
\(\Leftrightarrow\left(a+c\right)\left(ab+b^2+bc+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
\(\circledast Với:a=-b\) , ta có :
\(P=\left(-b+b\right)\left(b^3+c^3\right)\left(c^5+a^5\right)=0\)
\(\circledast Với:b=-c\) , ta có :
\(P=\left(a+b\right)\left(b^3-b^3\right)\left(c^5+a^5\right)=0\)
\(\circledast Với:c=-a\) , ta có :
\(P=\left(a+b\right)\left(b^3+c^3\right)\left(-a^5+a^5\right)=0\)
KL..............