Để cho dễ nhìn, đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x;y;z\right)\Rightarrow xyz=1\)
\(P=\left(\frac{y^2}{x}+\frac{z^2}{y}+\frac{x^2}{z}\right)+\left(\frac{z^2}{x}+\frac{x^2}{y}+\frac{y^2}{z}\right)\)
\(P\ge\frac{\left(x+y+z\right)^2}{x+y+z}+\frac{\left(x+y+z\right)^2}{x+y+z}=2\left(x+y+z\right)\ge2.3\sqrt[3]{xyz}=6\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;1\right)\) hay \(\left(a;b;c\right)=\left(1;1;1\right)\)
Nguyễn Việt Lâm, @Nk>t@ help me
Cách khác:
\(abc=1\)=> \(a=\frac{1}{bc}\) =>\(\sqrt{a}=\frac{1}{\sqrt{bc}}\)
Tương tự=> \(\sqrt{b}=\frac{1}{\sqrt{ac}}\),\(\sqrt{c}=\frac{1}{\sqrt{ab}}\)
Có \(\frac{b+c}{\sqrt{a}}+\frac{a+c}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\ge2\left(\frac{\sqrt{bc}}{\frac{1}{\sqrt{bc}}}+\frac{\sqrt{ac}}{\frac{1}{\sqrt{ac}}}+\frac{\sqrt{ab}}{\frac{1}{\sqrt{ab}}}\right)=2\left(bc+ac+ab\right)\ge2.3\sqrt[3]{\left(abc\right)^2}=2.3\sqrt[3]{1}=6\)
( theo bđt cosi)
Dấu "=" xảy ra <=>a=b=c=1
