\(\sqrt[3]{9}T=\sqrt[3]{3}\Sigma\sqrt{a+2b}=\Sigma\sqrt[3]{3.3\left(a+2b\right)}\le\dfrac{a+2b+3+3+b+2c+3+3+c+2a+3+3}{3}=\dfrac{3\left(a+b+c\right)+18}{3}=9\Leftrightarrow T\le\dfrac{9}{\sqrt[3]{9}}\)
\(2,,,,K=\Sigma\sqrt{x^2+\dfrac{1}{x^2}}\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x^{ }}+\dfrac{1}{y^{ }}+\dfrac{1}{z}\right)^2}\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{9}{x+y+z}\right)^2}=\sqrt{\left(x+y+z\right)^2+\dfrac{81}{\left(x+y+z\right)^2}}=\sqrt{\left(x+y+z\right)^2+\dfrac{1}{\left(x+y+z\right)^2}+\dfrac{80}{\left(x+y+z\right)^2}}\ge\sqrt{2+80}=\sqrt{82}\)
1) \(T=\sqrt[3]{a+2b}+\sqrt[3]{b+2c}+\sqrt[3]{c+2a}\)
\(=\dfrac{1}{\sqrt[3]{9}}.\sqrt[3]{\left(a+2b\right).3.3}+\dfrac{1}{\sqrt[3]{9}}.\sqrt[3]{\left(b+2c\right).3.3}+\dfrac{1}{\sqrt[3]{9}}.\sqrt[3]{\left(c+2a\right).3.3}\)
\(\le\dfrac{1}{\sqrt[3]{9}}.\left[\dfrac{\left(a+2b\right)+3+3}{3}+\dfrac{\left(b+2c\right)+3+3}{3}+\dfrac{\left(c+2a\right)+3+3}{3}\right]\)
\(=\dfrac{1}{\sqrt[3]{9}}.\left[\left(a+b+c\right)+6\right]=\dfrac{1}{\sqrt[3]{9}}.\left(3+6\right)=\dfrac{9}{\sqrt[3]{9}}=3\sqrt[3]{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Vậy \(MaxT=3\sqrt[3]{3}\)
Mk ko dùng kí hiệu sigma cyc nhé (\(\sum\)), sợ bạn không hiểu.
2) \(K=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(=\dfrac{1}{\sqrt{1^2+9^2}}.\sqrt{\left(x^2+\dfrac{1}{x^2}\right)\left(1^2+9^2\right)}+\dfrac{1}{\sqrt{1^2+9^2}}.\sqrt{\left(y^2+\dfrac{1}{y^2}\right)\left(1^2+9^2\right)}+\dfrac{1}{\sqrt{1^2+9^2}}.\sqrt{\left(z^2+\dfrac{1}{z^2}\right)\left(1^2+9^2\right)}\)
\(\ge\dfrac{1}{\sqrt{82}}\left[\sqrt{\left(x.1+\dfrac{1}{x}.9\right)^2}+\sqrt{\left(y.1+\dfrac{1}{y}.9\right)^2}+\sqrt{\left(z.1+\dfrac{1}{z}.9\right)^2}\right]\)
\(=\dfrac{1}{\sqrt{82}}.\left[\left(x+\dfrac{9}{x}\right)+\left(y+\dfrac{9}{y}\right)+\left(z+\dfrac{9}{z}\right)\right]\)
\(=\dfrac{1}{\sqrt{82}}.\left[\left(x+\dfrac{1}{9x}\right)+\left(y+\dfrac{1}{9y}\right)+\left(z+\dfrac{1}{9z}\right)+\dfrac{80}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]\)
\(\ge\dfrac{1}{\sqrt{82}}\left(2\sqrt{x.\dfrac{1}{9x}}+2\sqrt{y.\dfrac{1}{9y}}+2\sqrt{z.\dfrac{1}{9z}}+\dfrac{80}{9}.\dfrac{9}{x+y+z}\right)\)
\(\ge\dfrac{1}{\sqrt{82}}.\left(\dfrac{2}{3}+\dfrac{2}{3}+\dfrac{2}{3}+\dfrac{80}{1}\right)\)
\(=\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Vậy \(MinK=\sqrt{82}\)
