Assume that \(a\ge b\ge c>0\),so \(a-c\ge0\).Multiply 2 sides of the inequality \(b\ge c\) by \(a-c\),we have : 

\(b\left(a-c\right)\ge c\left(a-c\right)\Leftrightarrow ab-bc+c^2\ge ac\Leftrightarrow\dfrac{b}{c}-\dfrac{b}{a}+\dfrac{c}{a}\ge1\left(1\right)\)

We have : \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\left(2\right)\)(AM-GM inequality)

\(\left(\dfrac{a}{b}-1\right)^2+\left(\dfrac{b}{c}-1\right)^2+\left(\dfrac{c}{a}-1\right)^2\ge0\left(3\right)\)

The inequalities (1),(2),(3) are added up to : \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a^2}{b^2}-\dfrac{2a}{b}+1+\dfrac{b^2}{c^2}-\dfrac{2b}{c}+1+\dfrac{c^2}{a^2}-\dfrac{2c}{a}+1\ge3\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{a}{b}-\dfrac{b}{c}-\dfrac{c}{a}\ge0\)

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)

Assume that a≥b≥c>0,so a−c≥0.Multiply 2 sides of the inequality b≥c by a−c

,we have : 

b(a−c)≥c(a−c)⇔ab−bc+c2≥ac⇔bc−ba+ca≥1(1)

We have : ab+ba≥2√ab.ba=2(2)

(AM-GM inequality)

(ab−1)2+(bc−1)2+(ca−1)2≥0(3)

The inequalities (1),(2),(3) are added up to : ab+bc+ca+a2b2−2ab+1+b2c2−2bc+1+c2a2−2ca+1≥3

⇔a2b2+b2c2+c2a2−ab−bc−ca≥0

⇔a2b2+b2c2+c2a2≥ab+bc+ca