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Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có :
\(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{12}\)
Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)
Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z^2\right)}\end{cases}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{2}{3}\)
Ta có : \(ab+bc+ca=2abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\P=\frac{x^3}{\left(2-x\right)^2}\end{cases}+\frac{y^3}{\left(2-y\right)^3}+\frac{z^3}{\left(2-z\right)^2}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\frac{x^3}{\left(2-x\right)^2}+\frac{2-x}{8}+\frac{2-x}{8}\ge3\sqrt[3]{\frac{x^3}{64}}=\frac{3x}{4}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(2-y\right)^2}+\frac{2-y}{8}+\frac{2-y}{8}\ge\frac{3y}{4}\\\frac{z^3}{\left(2-z\right)^2}+\frac{2-z}{8}+\frac{2-z}{8}\ge\frac{3z}{8}\end{cases}}\)
\(\Rightarrow P+\frac{12-2\left(x+y+z\right)}{8}\ge\frac{3}{4}\left(x+y+z\right)\)
\(\Rightarrow P\ge\frac{1}{2}\)
Dấu " = " xảy ra khi \(x=y=z=\frac{2}{3}\)
Ta có : \(\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(ab+a+b+1\right)\left(c+1\right)\)
\(=abc+ab+ac+a+bc+b+c+1\)
\(=abc+ab+bc+ca+a+b+c+1\)
\(=\left(ab+bc+ca+abc\right)+\left(a+b+c\right)+1\)
Gỉa thiết tương đương với \(\left(ab+bc+ca+abc\right)+\left(a+b+c+1\right)=2+\left(a+b+c+1\right)\)
\(< =>\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(a+1\right)+\left(b+1\right)+\left(c+1\right)\)
\(< =>\frac{1}{\left(a+1\right)\left(b+1\right)}+\frac{1}{\left(b+1\right)\left(c+1\right)}+\frac{1}{\left(c+1\right)\left(a+1\right)}=1\)
Đặt \(\left(\frac{1}{a+1};\frac{1}{b+1};\frac{1}{c+1}\right)\rightarrow\left(x;y;z\right)\)(x;y;z>0) Thì bài toán quy về :
Biết \(xy+yz+zx=1\)
Tìm Max của \(M=\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)
Có \(x^2+1=x^2+xy+yz+zx=x\left(x+y\right)+z\left(x+y\right)=\left(x+z\right)\left(x+y\right)\)
Tương tự \(y^2+1=\left(y+x\right)\left(y+z\right);z^2+1=\left(z+x\right)\left(z+y\right)\)
Suy ra \(M=\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+x\right)\left(y+z\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\)
\(=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Ta có bổ đề quen thuộc sau : \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)(link chứng minh mình đặt ở cuối bài)
\(< =>\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\frac{8}{9}\left(x+y+z\right)\)(*)
Có \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3< =>x+y+z\ge\sqrt{3}\)
Suy ra (*) tương đương \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge\frac{8}{9}.\sqrt{3}=\frac{8\sqrt{3}}{9}\)
Khi đó \(M\le\frac{2}{\frac{8\sqrt{3}}{9}}=\frac{2.9}{8\sqrt{3}}=\frac{18}{8\sqrt{3}}\)
p/s : cách chứng minh bổ đề trên đây ạ : https://olm.vn/hoi-dap/detail/89229441423.html
\(\Leftrightarrow\Sigma_{cyc}\frac{\left(ab+bc+ca\right)^2}{2a^2+bc}\le\left(a+b+c\right)^2\)
Ta có: \(\frac{\left(ab+bc+ca\right)^2}{2a^2+bc}\le\frac{\left(ab+ca\right)^2}{2a^2}+\frac{\left(bc\right)^2}{bc}=\frac{\left(b+c\right)^2}{2}+bc\)
Tương tự rồi cộng lại ta thu được:
\(L.H.S\le\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{2}+ab+bc+ca\)
\(=\frac{2\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)}{2}+ab+bc+ca\)\(=\left(a+b+c\right)^2\)
P/s: Nhìn đơn giản chứ nó là bao nhiêu ngày suy nghĩ đấy ạ:( Chả biết đúng hay sai nữa:v
Bài 2:
\(\frac{1}{\sqrt[3]{81}}\cdot P=\frac{1}{\sqrt[3]{9\cdot9\cdot\left(a+2b\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(b+2c\right)}}+\frac{1}{\sqrt[3]{9\cdot9\cdot\left(c+2a\right)}}\)
\(\ge\frac{3}{a+2b+9+9}+\frac{3}{b+2c+9+9}+\frac{3}{c+2a+9+9}\ge3\left(\frac{9}{3a+3b+3c+54}\right)=\frac{1}{3}\)
\(\Rightarrow P\ge\sqrt[3]{3}\)
Dấu bằng xẩy ra khi a=b=c=3
Bài 1:
\(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)
Theo bđt côsi-shaw ta luôn có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge\frac{25}{x+y+z+t+k}\)(x=y=z=t=k>0 ) (*)
\(\Leftrightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}x+y+z+t+k\ge5\sqrt[5]{xyztk}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\ge5\sqrt[5]{\frac{1}{xyztk}}\end{cases}}\)
\(\Rightarrow\left(x+y+z+t+k\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\ge25\)
\(\Rightarrow\)(*) luôn đúng
Từ (*) \(\Rightarrow\frac{1}{25}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}+\frac{1}{k}\right)\le\frac{1}{x+y+z+t+k}\)
Ta có: \(P=\frac{1}{2a+2b+c}+\frac{1}{a+2b+2c}+\frac{1}{2a+b+2c}\)
Mà \(\frac{1}{2a+2b+c}=\frac{1}{a+a+b+b+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\frac{1}{a+2b+2c}=\frac{1}{a+b+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\frac{1}{2a+b+2c}=\frac{1}{a+a+b+c+c}\le\frac{1}{25}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{25}\left[5.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right]=1\)
\(\Rightarrow P\le1\left(đpcm\right)\)Dấu"="xảy ra khi a=b=c\(=\frac{3}{5}\)
Dùng Cô-si ngược dấu:
Ta có : a\(1+b^2)=a-(ab^2/(1+b^2))>=a-(ab^2/2b)=...
Tương tự ta có:b/(1+c^2)>=b-bc/2
c/(1+a^2)>=c-ac/2
Cộng vế với vế ta có A>=(a+b+c)-(ab+bc+ca)/2
Mà 3(ab+bc+ca)<=a^2+b^2+c^2+2ab+2bc+2ca
<=>3(ab+bc+ca)<=(a+b+c)^2
<=>-(ab+bc+ca)>=-(a+b+c)^2/3
Thay vào ta có: A>=(a+b+c)-(a+b+c)^2/6=3/2
Dấu = xảy ra<=>a=b=c=1/3
đề bài của mình mẫu là 1+2b^2 ko phải 1+b^2