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Theo Svac - xơ có :
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\)
Khi đó \(P\ge\frac{9}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2}\)
\(=\left(\frac{1}{ab+bc+ca}+\frac{1}{ab+bc+ca}+\frac{1}{a^2+b^2+c^2}\right)+\frac{7}{ab+bc+ca}\)
\(\ge\frac{9}{a^2+b^2+c^2+2.\left(ab+bc+ca\right)}+\frac{7}{\frac{\left(a+b+c\right)^2}{3}}\)
\(=\frac{9}{\left(a+b+c\right)^2}+\frac{21}{\left(a+b+c\right)^2}=\frac{30}{\left(a+b+c\right)^2}=\frac{10}{3}\)
Dấu "=: xảy ra khi \(a=b=c=1\)
Vậy \(P_{min}=\frac{10}{3}\) khi \(a=b=c=1\)
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 \(x=y=z=\frac{2}{3}\)
Do a, b, c dương áp dụng bất đẳng thức Cô-si ta có:
\(\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}\ge2\sqrt{\frac{b^2c^2}{a^2}.\frac{a^2c^2}{b^2}}=2c^2\)(1)
Tương tự \(\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}\ge2a^2\) (2) và \(\frac{b^2c^2}{a^2}+\frac{a^2b^2}{c^2}\ge2b^2\) (3)
Cộng (1), (2), (3) vế theo vế rồi chia 2 vế cho 2 ta được \(\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}\ge a^2+b^2+c^2=1\)
Ta có \(P^2=\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}+2\left(\frac{bc}{a}.\frac{ac}{b}+\frac{ac}{b}.\frac{ab}{c}+\frac{bc}{a}.\frac{ab}{c}\right)\)
\(P^2=\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}+2\left(a^2+b^2+c^2\right)=\frac{b^2c^2}{a^2}+\frac{a^2c^2}{b^2}+\frac{a^2b^2}{c^2}+2\ge1+2=3\)
Vậy \(P_{min}=\sqrt{3}\) \(\Leftrightarrow\) \(a=b=c=\frac{\sqrt{3}}{3}\)
Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\) (áp dụng Bất Đẳng Thức Cosi)
\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)
\(\Rightarrow\frac{a^2}{b}-a+2b\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\left(1\right)\)
Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)
Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)
Dấu "=" xảy ra khi a=b=c
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 \(x=y=z=\frac{2}{3}\)
GT => (a+1)(b+1)(c+1)=(a+1)+(b+1)+(c+1)
Đặt \(\frac{1}{a+1}=x,\frac{1}{1+b}=y,\frac{1}{c+1}=z\), ta cần tìm min của\(\frac{x}{x^2+1}+\frac{y}{y^2+1}+\frac{z}{z^2+1}\)với xy+yz+zx=1
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\Leftrightarrow\frac{2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Mà (x+y)(y+z)(z+x) >= 8/9 (x+y+z)(xy+yz+xz) >= \(\frac{8\sqrt{3}}{9}\) nên \(M\)=< \(\frac{3\sqrt{3}}{4}\),dấu bằng xảy ra khi a=b=c=\(\sqrt{3}-1\)
Theo giả thiết, ta có: \(abc+ab+bc+ca=2\)
\(\Leftrightarrow abc+ab+bc+ca+a+b+c+1=a+b+c+3\)
\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)=\left(a+1\right)+\left(b+1\right)+\left(c+1\right)\)
\(\Leftrightarrow\frac{1}{\left(a+1\right)\left(b+1\right)}+\frac{1}{\left(b+1\right)\left(c+1\right)}+\frac{1}{\left(c+1\right)\left(a+1\right)}=1\)
Đặt \(\left(a+1;b+1;c+1\right)\rightarrow\left(\frac{\sqrt{3}}{x};\frac{\sqrt{3}}{y};\frac{\sqrt{3}}{z}\right)\). Khi đó giả thiết bài toán được viết lại thành xy + yz + zx = 3
Ta có: \(M=\Sigma_{cyc}\frac{a+1}{a^2+2a+2}=\Sigma_{cyc}\frac{a+1}{\left(a+1\right)^2+1}\)\(=\Sigma_{cyc}\frac{1}{a+1+\frac{1}{a+1}}=\Sigma_{cyc}\frac{1}{\frac{\sqrt{3}}{x}+\frac{x}{\sqrt{3}}}\)
\(=\sqrt{3}\left(\frac{x}{x^2+3}+\frac{y}{y^2+3}+\frac{z}{z^2+3}\right)\)
\(=\sqrt{3}\text{}\Sigma_{cyc}\left(\frac{x}{x^2+xy+yz+zx}\right)=\sqrt{3}\Sigma_{cyc}\frac{x}{\left(x+y\right)\left(x+z\right)}\)
\(\le\frac{\sqrt{3}}{4}\Sigma_{cyc}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)=\frac{3\sqrt{3}}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)hay \(a=b=c=\sqrt{3}-1\)
\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{a^2+b^2+c^2}\)
\(P=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{9-2\left(ab+bc+ca\right)}\)
\(P=\frac{1}{3ab}+\frac{1}{3bc}+\frac{1}{3ca}+\frac{1}{9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(P\ge\frac{16}{3ab+3bc+3ca+9-2\left(ab+bc+ca\right)}+\frac{2}{3}\left(\frac{9}{ab+bc+ca}\right)\)
\(P\ge\frac{16}{9+ab+bc+ca}+\frac{6}{ab+bc+ca}\)
Sử dụng đánh giá quen thuộc:\(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\)
\(\Rightarrow ab+bc+ca\le3\)
\(\Rightarrow P\ge\frac{16}{9+3}+\frac{6}{3}=2+\frac{4}{3}=\frac{10}{3}\)
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