Applying the Cauchyschwarz inequality to the Engel form we have:

\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ca}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge\dfrac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

We have things to prove

Apply AM-GM inequality for 2 numbers , we have :

\(\dfrac{a^3}{b}+ab\ge2.\sqrt{\dfrac{a^3}{b}.ab}=2a^2\)

\(\dfrac{b^3}{c}+bc\ge2.\sqrt{\dfrac{b^3}{c}.bc}=2b^2\)

\(\dfrac{c^3}{a}+ca\ge2\sqrt{\dfrac{c^3}{a}.ca}=2c^2\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2a^2+2b^2+2c^2\)

In another case , we have this :

\(a^2+b^2+c^2\ge ab+bc+ca\) (The consequence of AM-GM)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}+ab+bc+ca\ge2ab+2bc+2ca\)

\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)

Apply AM-GM inequality for 2 numbers , we have :

a3b+ab≥2.√a3b.ab=2a2

b3c+bc≥2.√b3c.bc=2b2

c3a+ca≥2√c3a.ca=2c2

⇒a3b+b3c+c3a+ab+bc+ca≥2a2+2b2+2c2

In another case , we have this :

a2+b2+c2≥ab+bc+ca

 (The consequence of AM-GM)

⇔2a2+2b2+2c2≥2ab+2bc+2ca

⇒a3b+b3c+c3a+ab+bc+ca≥2ab+2bc+2ca

⇒a3b+b3c+c3a≥ab+bc+ca

I choose Lufans's answer because it's simple.

a,b>0; c<0 ? or c=0 ? seem to have a typo