\(A^2=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2\left(b^2+c^2+a^2\right)=\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}+2\)
Áp dụng Côsi: \(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}\ge2\sqrt{\frac{a^2b^2}{c^2}.\frac{b^2c^2}{a^2}}=2\sqrt{b^4}=2b^2\)
Tương tự \(\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\ge2c^2;\text{ }\frac{c^2a^2}{b^2}+\frac{a^2b^2}{c^2}\ge2a^2\)
\(\Rightarrow2\left(\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\right)\ge2\left(a^2+b^2+c^2\right)=2\)
\(\Rightarrow\frac{a^2b^2}{c^2}+\frac{b^2c^2}{a^2}+\frac{c^2a^2}{b^2}\ge1\)
\(\Rightarrow A^2\ge1+2=3\)
\(\Rightarrow A\ge\sqrt{3}\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{\sqrt{3}}\)
