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Ta có: \(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Rightarrow\frac{a}{b-c}=-\frac{b}{c-a}-\frac{c}{a-b}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-b\left(a-b\right)-c\left(c-a\right)}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{b-c}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)}\)
\(\Rightarrow\frac{a}{\left(b-c\right)^2}=\frac{-ab+b^2-c^2+ac}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Tương tự ta có: \(\frac{b}{\left(c-a\right)^2}=\frac{-bc+c^2-a^2+ab}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
\(\frac{c}{\left(a-b\right)^2}=\frac{-ca+a^2-b^2+bc}{\left(c-a\right)\left(a-b\right)\left(b-c\right)}\)
Cộng các đẳng thức trên ta được:
\(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)\(\frac{-ab+b^2-c^2+ac-bc+c^2-a^2+ba-ca+a^2-b^2+bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Vậy \(\frac{a}{\left(b-c\right)^2}\)\(+\frac{b}{\left(c-a\right)^2}\)\(+\frac{c}{\left(a-b\right)^2}=\)0 (đpcm)
Bài 1: diendantoanhoc.net
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\) BĐT cần chứng minh trở thành
\(\frac{x}{\sqrt{3zx+2yz}}+\frac{x}{\sqrt{3xy+2xz}}+\frac{x}{\sqrt{3yz+2xy}}\ge\frac{3}{\sqrt{5}}\)
\(\Leftrightarrow\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}+\frac{y}{\sqrt{5x}\cdot\sqrt{3y+2z}}+\frac{z}{\sqrt{5y}\cdot\sqrt{3z+2x}}\ge\frac{3}{5}\)
Theo BĐT AM-GM và Cauchy-Schwarz ta có:
\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{x}{\sqrt{5z}\cdot\sqrt{3x+2y}}\ge2\)\( {\displaystyle \displaystyle \sum }\)\(\frac{x}{3x+2y+5z}\ge\frac{2\left(x+y+z\right)^2}{x\left(3x+2y+5z\right)+y\left(5x+3y+2z\right)+z\left(2x+5y+3z\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+7\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(xy+yz+zx\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{3\left(x^2+y^2+z^2\right)+\frac{1}{3}\left(x^2+y^2+z^2\right)+\frac{20}{3}\left(xy+yz+zx\right)}\)
\(=\frac{2\left(x^2+y^2+z^2\right)}{5\left[x^2+y^2+z^2+2\left(xy+yz+zx\right)\right]}=\frac{3}{5}\)
Bổ sung bài 1:
BĐT được chứng minh
Đẳng thức xảy ra <=> a=b=c
ta có a/b+c+b/a+c+c/a+b=1
=> (a+b+c)(a/b+c+b/a+c+c/a+b)=a+b+c
=> a^2/b+c+ab/a+c+ac/a+b+ba/b+c+b^2/a+c+bc/a+b+ca/b+c+bc/a+c+c^2/a+b=a+b+c
=> a^2/b+c+(ba/b+c+ca/b+c)+b^2/a+c+(ab/a+c+bc/a+c)+c^2/a+b+(ac/a+b+bc/a+b)=a+b+c
=>( a^2/b+c)+a+(b^2/a+c)+b+(c^2/a+b)+c=a+b+c
=> a^2/b+c+b^2/a+c+c^2/a+b=0
Lời giải ở đây nhé:
http://olm.vn/hoi-dap/question/415281.html
Xét: \(9M=\Sigma\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{3}{2}+\Sigma\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-3+\frac{9}{2}\)
\(=\Sigma\left(\frac{a^2+b^2+c^2}{4a^2+b^2+c^2}-\frac{1}{2}\right)+\Sigma\left(\frac{2\left(ab+bc+ca\right)}{4a^2+b^2+c^2}-1\right)+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{b^2+c^2-2a^2}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2ab+2bc+2ca-4a^2-b^2-c^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\frac{\left(b-a\right)\left(b+a\right)+\left(c-a\right)\left(c+a\right)}{\left(4a^2+b^2+c^2\right)}+\Sigma\frac{2a\left[\left(b-a\right)+\left(c-a\right)\right]}{4a^2+b^2+c^2}-\Sigma\frac{\left(b-c\right)^2}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(\frac{\left(a-b\right)\left(a+b\right)}{a^2+4b^2+c^2}-\frac{\left(a-b\right)\left(b+a\right)}{4a^2+b^2+c^2}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\frac{1}{2}\Sigma\left(a-b\right)\left(a+b\right)\left(\frac{3a^2-3b^2}{\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}\right)-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\frac{3\left(a-b\right)^2\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}-\Sigma\frac{\left(a-b\right)^2}{a^2+b^2+4c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}-\frac{1}{a^2+b^2+4c^2}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)
\(=\Sigma\left(a-b\right)^2\left[\frac{3\left(a+b\right)^2\left(a^2+b^2+4c^2\right)-2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)}{2\left(a^2+4b^2+c^2\right)\left(4a^2+b^2+c^2\right)\left(a^2+b^2+4c^2\right)}\right]-\Sigma\frac{2a\left(a-b\right)}{4a^2+b^2+c^2}+\frac{9}{2}\)Ai đó làm tiếp giúp em vs:( Em chỉ nghĩ ra được tới đây thôi.
Ta có:
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab;a^2+c^2\ge2\sqrt{a^2c^2}=2ac;a^2+a^2\ge2\sqrt{a^2a^2}=2a^2\)
Khi đó:
\(4a^2+b^2+c^2\ge2a\left(a+b+c\right)\)
\(\Rightarrow\frac{1}{4a^2+b^2+c^2}\le\frac{1}{6a}\)
Tương tự:
\(\frac{1}{a^2+4b^2+c^2}\le\frac{1}{6b};\frac{1}{a^2+b^2+4c^2}\le\frac{1}{6c}\cdot\)
\(\Rightarrow M\le\frac{1}{6}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+bc+ca}{abc}\cdot\frac{1}{6}\) \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow3\ge3\sqrt[3]{abc}\Rightarrow abc\le1\)
Theo BĐT \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Khi đó \(M\le\frac{3}{1}\cdot\frac{1}{6}=\frac{1}{2}\)
Dấu "=" xảy ra tại \(a=b=c=1\)
P/S:Is that true ??
1) \(21x^2+21y^2+z^2\)
\(=18\left(x^2+y^2\right)+z^2+3\left(x^2+y^2\right)\)
\(\ge9\left(x+y\right)^2+z^2+3.2xy\)
\(\ge2.3\left(x+y\right).z+6xy\)
\(=6\left(xy+yz+zx\right)=6.13=78\)
Dấu "=" xảy ra <=> x = y ; 3(x+y) = z; xy + yz + zx= 13 <=> x = y = 1; z= 6
2) \(x+y+z=3xyz\)
<=> \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=3\)
Đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)=> ab + bc + ca = 3
Ta cần chứng minh: \(3a^2+b^2+3c^2\ge6\)
Ta có: \(3a^2+b^2+3c^2=\left(a^2+c^2\right)+2\left(a^2+c^2\right)+b^2\)
\(\ge2ac+\left(a+c\right)^2+b^2\ge2ac+2\left(a+c\right).b=2\left(ac+ab+bc\right)=6\)
Vậy: \(\frac{3}{x^2}+\frac{1}{y^2}+\frac{3}{z^2}\ge6\)
Dấu "=" xảy ra <=> a = c = \(\sqrt{\frac{3}{5}}\); \(b=2\sqrt{\frac{3}{5}}\)
khi đó: \(x=z=\sqrt{\frac{5}{3}};y=\sqrt{\frac{5}{3}}\)
a) Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có :
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\)
Tương tự :
\(\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{4}{b+c-a+c+a-b}=\frac{4}{2c}=\frac{2}{c}\)
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{4}{a+b-c+c+a-b}=\frac{4}{2a}=\frac{2}{a}\)
Cộng theo vế :
\(\Rightarrow2\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(đpcm\right)\)
Đẳng thức xảy ra <=> a = b = c
\(\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\)
\(=\frac{a^4}{ab+ac}+\frac{b^4}{cb+ba}+\frac{c^4}{ac+bc}\)
\(\ge\frac{\left(a^2+b^2+c\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{2\left(ab+bc+ca\right)}\)
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
bài này chắc có câu a đúng ko
ta có \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=\frac{a}{c}=\frac{c}{b}=\frac{b}{a}\)
\(\Leftrightarrow a^4c^2+b^4a^2+c^4b^2=abc\left(a^2c+c^2a+b^2c\right)\)
đặt \(x=a^2c;y=b^2a;z=c^2b\)ta được
\(x^2+y^2+z^2=xy+yz+zx\)
áp dụng kết quả của câu a ta đc
\(\left(x-y\right)^2+\left(y-2\right)^2+\left(z-x\right)^2=0=>x=y=z\)
\(=>a^2c=b^2a=c^2b=>ac=b^2;bc=a^2;ab=c^2\)
=>a=b=c(dpcm)
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}=\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\)
Đặt \(\frac{a}{b}=x;\frac{b}{c}=y;\frac{c}{a}=z\)
Khi đó:\(x^2+y^2+z^2=xy+yz+zx\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)=2\left(xy+yz+zx\right)\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Mà \(\left(x-y\right)^2\ge0;\left(y-z\right)^2\ge0;\left(z-x\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
Dấu "=" xảy ra tại x=y=z hay a=b=c
Suy ra điều fải chứng minh