We have: \(a^2+b^2+c^2\ge ab+bc+ca\)

[\(a^2+b^2+c^2\ge ab+bc+ca\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+ac+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2\)

\(\ge0\) (proved)]

=> \(a^2+b^2+c^2+2ab+2bc+2ca\ge ab+bc+ca\)

=> \(\left(a+b+c\right)^2\ge3\left(ab+ac+ca\right)\)

=> \(\left(a+b+c\right)^3\ge3\left(ab+ac+ca\right)\) (proved)

We have: a2+b2+c2≥ab+bc+ca

[a2+b2+c2≥ab+bc+ca⇔2(a2+b2+c2)≥2(ab+ac+ca)⇔(a−b)2+(b−c)2

≥0

 (proved)]

=> a2+b2+c2+2ab+2bc+2ca≥ab+bc+ca

=> (a+b+c)2≥3(ab+ac+ca)

=> (a+b+c)3≥3(ab+ac+ca)

 (proved)