cho a,b,c là các số thực dương thỏa mãn abc=1 Tìm GTLN của \(P=\frac{1}{a+2b+3}+\frac{1}{b+2c+3}+\frac{1}{c+2a+3}\)
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Lời giải :
\(P=\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\)
\(P=\frac{1}{9}\cdot\left(\frac{9}{a+b+b}+\frac{9}{b+c+c}+\frac{9}{c+a+a}\right)\)
Áp dụng bđt Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)ta có :
\(P\le\frac{1}{9}\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{b}+\frac{2}{c}+\frac{1}{c}+\frac{2}{a}\right)\)
\(=\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{1}{3}\cdot9=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Theo Cauchy: \(\frac{1}{a+2b}=\frac{1}{a+b+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3\)
Đẳng thức xảy ra khi a = b = c = 1.
Vậy..
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Áp dụng bổ đề quen thuộc \(x^3+y^3\ge xy\left(x+y\right)\), ta được: \(\frac{1}{2a^3+b^3+c^3+2}=\frac{1}{\left(a^3+b^3\right)+\left(a^3+c^3\right)+2}\le\frac{1}{ab\left(a+b\right)+ac\left(a+c\right)+2}\)\(=\frac{bc}{ab^2c\left(a+b\right)+abc^2\left(a+c\right)+2bc}=\frac{bc}{b\left(a+b\right)+c\left(a+c\right)+2bc}\)\(\le\frac{bc}{ab+ac+4bc}=\frac{bc}{b\left(a+c\right)+c\left(a+b\right)+2bc}\)\(\le\frac{1}{9}\left(\frac{bc}{b\left(a+c\right)}+\frac{bc}{c\left(a+b\right)}+\frac{bc}{2bc}\right)=\frac{1}{9}\left(\frac{c}{a+c}+\frac{b}{a+b}+\frac{1}{2}\right)\)(1)
Tương tự, ta có: \(\frac{1}{a^3+2b^3+c^3+2}\le\frac{1}{9}\left(\frac{c}{b+c}+\frac{a}{a+b}+\frac{1}{2}\right)\)(2); \(\frac{1}{a^3+b^3+2c^3+2}\le\frac{1}{9}\left(\frac{b}{b+c}+\frac{a}{a+c}+\frac{1}{2}\right)\)(3)
Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(P\le\frac{1}{9}\left(1+1+1+\frac{3}{2}\right)=\frac{1}{2}\)
Vậy giá trị lớn nhất của P là \(\frac{1}{2}\)đạt được khi x = y = z = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2ab+2b+2}=\frac{1}{2}\cdot\frac{1}{ab+b+1}\)
Tương tự CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1}\) và \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)=\frac{1}{2}\cdot1=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
A=\(\frac{1}{a^2+2b^2+3}\)+\(\frac{1}{b^2+2c^2+3}\)+\(\frac{1}{c^2+2a^2+3}\)
ta có: \(\frac{1}{a^2+2b^2+3}\)=\(\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\)\(\le\)\(\frac{1}{2\left(ab+b+1\right)}\)
vì : a2+b2\(\ge\)2\(\sqrt{a^2b^2}\)=2ab
b2+1\(\ge\)2\(\sqrt{b^2x1}\)=2b
cmtt => A\(\le\)\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{1}{bc+c+1}\)+\(\frac{1}{ca+a+1}\))
=\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab^2c+abc+ab}\)+\(\frac{b}{cba+ab+b}\))
=\(\frac{1}{2}\)x (\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab+b+1}\)+\(\frac{b}{ab+b+1}\))=\(\frac{1}{2}\)x\(\frac{ab+b+1}{ab+b+1}\)=\(\frac{1}{2}\)
dấu "=" xảy ra <=> a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{1}{\left(a+2\right)+\left(a+2\right)+\left(b+2\right)}+\frac{1}{\left(b+2\right)+\left(b+2\right)+\left(c+2\right)}+\frac{1}{\left(c+2\right)+\left(c+2\right)+\left(a+2\right)}\)
\(\le\frac{1}{9}\left(\frac{2}{a+2}+\frac{1}{b+2}\right)+\frac{1}{9}\left(\frac{2}{b+2}+\frac{1}{c+2}\right)+\frac{1}{9}\left(\frac{2}{c+2}+\frac{1}{a+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}\right)\)
Dễ dàng cm BĐT \(\frac{1}{x+1}+\frac{1}{y+1}\ge\frac{2}{1+\sqrt{xy}}\)
\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=\frac{1}{2}\left(\frac{1}{1+\frac{a}{2}}+\frac{1}{1+\frac{b}{2}}+\frac{1}{1+\frac{c}{2}}\right)\)
\(\le\frac{1}{2}.\frac{3}{1+\sqrt[3]{\frac{abc}{8}}}=\frac{3}{4}\Rightarrow P\le\frac{1}{4}\)
Xảy ra khi \(a=b=c=2\)
À viết ngược dấu BĐT phụ r` :v
\(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\) mới đúng nhé :v
\(\Leftrightarrow\frac{\left(\sqrt{xy}-1\right)\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x+1\right)\left(y+1\right)\left(1+\sqrt{xy}\right)}\le0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(p=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\)
Áp dụng bất đẳng thức AM - GM ta có :
\(\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4ab}}=\frac{1}{a}\)
\(\frac{ac}{b^2\left(a+c\right)}+\frac{a+c}{4ac}\ge4\sqrt{\frac{ac}{b^2\left(a+c\right)}.\frac{a+c}{4ac}}=\frac{1}{b}\)
\(\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}.\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng vế với vế ta được \(p+\frac{1}{4c}+\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4a}+\frac{1}{4c}+\frac{1}{4b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow p+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow p\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{2a.2b.2c}}=\frac{3}{\sqrt[3]{8abc}}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Xét: \(\frac{bc}{a^2b+ca^2}=\frac{bc}{a\cdot abc\cdot\frac{1}{c}+a\cdot abc\cdot\frac{1}{b}}=\frac{b^2c^2}{ab+ca}\)(*)
Tương tự với (*) ta có: \(\hept{\begin{cases}\frac{ca}{b^2c+ab^2}=\frac{c^2a^2}{ab+bc}\\\frac{ab}{c^2a+bc^2}=\frac{a^2b^2}{ca+bc}\end{cases}}\)
\(\Rightarrow\Sigma_{cyc}\frac{bc}{a^2b+ca^2}=\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\)
Ta thấy\(\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\) có dạng: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\left(a+b+c\right)\)
Bước cuối Cô-si ba số và kết hợp điều kiện abc=1 là xong
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(5a^2+2ab+2b^2=4a^2+2ab+b^2+\left(a^2+b^2\right)\ge4a^2+2ab+b^2+2ab=\left(2a+b\right)^2\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)
Lại có: \(\frac{1}{2a+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{9}\left(\frac{2}{a}+\frac{1}{b}\right)\)
Tương tự cộng lại ta có: \(VT\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Theo BĐT Bunhiacopxki ta có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=3\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le\sqrt{3}\)
\(\Rightarrow VT\le\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\)
Dấu = xảy ra khi \(a=b=c=\sqrt{3}\)
Đặt \(a=x^2;b=y^2;c=z^2\)khi đó ta được xyz=1 và biểu thức P viết được thành
\(P=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2x^2+3}+\frac{1}{z^2+2x^2+3}\)
Ta có \(x^2+y^2\ge2xy;y^2+1\ge2y\Rightarrow x^2+2y^2+3\ge2\left(xy+y+1\right)\)
Do đó ta được \(\frac{1}{x^2+2y^2+3}\le\frac{1}{2}\cdot\frac{1}{xy+y+1}\)
Chứng minh tương tự ta có:
\(\frac{1}{y^2+2z^2+3}\le\frac{1}{2}\cdot\frac{1}{yz+z+1};\frac{1}{z^2+2x^2+3}\le\frac{1}{2}\cdot\frac{1}{zx+z+1}\)
Cộng các vế BĐT trên ta được
\(P\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
Ta cần chứng minh \(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+a+1}=1\)
Do xyz=1 nên ta được
\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{zx}{z+1+zx}+\frac{x}{1+zx+z}+\frac{1}{zx+x+1}=1\)
Từ đó ta được
\(P\le\frac{1}{2}\). Dấu "=" xảy ra <=> a=b=c=1