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(a+b+c)2=a2+b2+c2
=>2(ab+bc+ac)=0
=>ab+bc+ac=0
=> bc=-ab-ac
=>\(\frac{a^2}{a^2+2bc}=\frac{a^2}{a^2-ac-ab+bc}\)=\(\frac{a^2}{\left(a-c\right)\left(a-b\right)}\)
Tuong tu => \(\frac{b^2}{b^2+2ac}=....\)
\(\frac{c^2}{c^2+2ab}=...\)
=> \(\frac{a^2}{a^2+2bc}+....\)=\(\frac{a^2}{\left(a-b\right)\left(a-c\right)}\)+...
=\(\frac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
=1
Ta có: \(A=a\left(a^2-bc\right)+b\left(b^2-ac\right)+c\left(c^2-ab\right)=0\)
\(\Rightarrow A=a^3+b^3+c^3-3abc=0\) \(\Rightarrow A=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow A=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-ac-bc=0\)
Xét \(M=a^2+b^2+c^2-ab-ac-bc=0\)
\(\Rightarrow2M=2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow2M=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Rightarrow a-b=0;b-c=0;c-a=0\) \(\Rightarrow a=b=c\)
\(\Rightarrow P=\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=1+1+1=3\)
Mình xem phép làm câu 1 ạ.
Đề là?
\(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\)(1)
Chứng minh tương đương
\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge4\)<=> 12ac - 9bc - 9ab + 6b2 \(\le\)0 ( quy đồng ) (2)
Từ (1) <=> 2ac = ab + bc Thay vào (2) <=> 6ab + 6bc - 9bc - 9ab + 6b2 \(\le\)0
<=> a + c \(\ge\)2b
Từ (1) => \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)
=> a + c \(\ge\)2b đúng => BĐT ban đầu đúng
Dấu "=" xảy ra <=> a = c = b
có \(a^3+b^3+c^3=3abc \Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)^3-3c\left(a+b\right)\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\end{cases}}\)
có \(S=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
mà \(a=b=c\left(cmt\right)\)
\(\Rightarrow S=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
a ) \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)
\(\Leftrightarrow a^2+b^2+c^2=0\)
Do \(a^2\ge0;b^2\ge0;c^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )
Thay * vào biểu thức M , ta được :
\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)
\(=-1^{1999}+0+1^{2001}\)
\(=-1+0+1\)
\(=0\)
Vậy \(M=0\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)
\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)
\(\Leftrightarrow bc+ac+ab-1=0\)
\(\Leftrightarrow bc+ac+ab=1\)
Mà \(a^2+b^2+c^2=1\)
\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)
\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)
\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)
\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)
\(\Rightarrow P=1+1+1=3\)
Vậy \(P=3\)
Em tham khảo cách làm tại link: Câu hỏi của Cao Chi Hieu - Toán lớp 9 - Học toán với OnlineMath
Ùi mình làm theo kiểu khác thử :V, nhưng có hơi hướng giống và bổ sung :D
Câu 2 : a,b,c > 0. CM : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Giải :
C1 : Áp dụng bất đẳng thức Cauchy - Schwarz dạng Engel ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\left(ĐPCM\right)\)
Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\).
C2 : Đầy đủ hơn với cách giải đúng của bạn Hoàng Thiên Di :
Áp dụng BĐT AM-GM cho 3 số dương (sgk là cosi :v)
\(a+b+c\ge3\sqrt[3]{abc}\)\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\ge3+2+2+2=9\left(ĐPCM\right)\)
Câu 3 : a,b,c > 0. CM : \(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
Giải :
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
\(\Leftrightarrow\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}\ge6\)
\(\Leftrightarrow\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge6\)
Theo bất đẳng thức Cosi : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{yx}}=2\)
Thay vào các vế được : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{ab}{ba}}=2\sqrt{1}=2\)
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{ac}{ca}}=2\sqrt{1}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{bc}{cb}}=2\sqrt{1}=2\)
\(\Leftrightarrow2+2+2\ge6\) (đúng)
BĐT được c/m.
1) \(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)
\(\left(do\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]=-\left(c^3-a^3\right)\right)\)
\(=\left(a-b\right)\left(b^3-c^3\right)+\left(c-b\right)\left(a^3-b^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c^2-a^2\right)+\left(bc-ab\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
2) \(\dfrac{a-b}{b+c}+\dfrac{b-a}{c+a}+\dfrac{c-b}{a+b}=1\)
\(\Rightarrow\dfrac{a-c}{b+c}+1+\dfrac{b-a}{c+a}+1+\dfrac{c-b}{a+b}+1=4\)
\(\Rightarrow\dfrac{a-c+b+c}{b+c}+\dfrac{b-a+c+a}{c+a}+\dfrac{c-b+a+b}{a+b}=4\)
\(\Rightarrow\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}=4\)
( ͡° ͜ʖ ͡° )
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