a: Ta có: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\cdot\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=2\cdot\sqrt{\frac{a^2}{c^2}}=2\cdot\frac{a}{c}\)
\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\cdot\sqrt{\frac{b^2}{c^2}\cdot\frac{c^2}{a^2}}=2\cdot\sqrt{\frac{b^2}{a^2}}=\frac{2b}{a}\)
\(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\cdot\sqrt{\frac{a^2}{b^2}\cdot\frac{c^2}{a^2}}=2\cdot\sqrt{\frac{c^2}{b^2}}=2\cdot\frac{c}{b}\)
Do đó: \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\)
=>\(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)\)
=>\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\)
b: \(a+b\ge2\cdot\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(a+c\ge2\sqrt{ac}\)
Do đó: \(\left(a+b\right)\left(b+c\right)\left(a+c\right)\ge2\cdot\sqrt{ab}\cdot2\sqrt{ac}\cdot2\sqrt{bc}=8\cdot\sqrt{a^2b^2c^2}=8abc\)
