We have:

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

Prove that it: 

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

<=> \(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

<=> \((a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0 \) (true) 

So a² + b² + c² \(\ge\) ab + bc + ca

Apply this for the topic, we have:

\(a^4+b^4+c^4 \ge (ab)^2 + (bc)^2 + (ca)^2 \)

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

We have:

a2+b2+c2≥ab+bc+ca

Prove that it: 

This inequality <=> 2a2+2b2+2c2≥2ab+2bc+2ca

<=> 2a2+2b2+2c2−2ab−2bc−2ca≥0

<=> (a−b)2+(b−c)2+(c−a)2≥0

 (true) 

So a2 + b2 + c2 ≥

 ab + bc + ca

Apply this for the topic, we have:

a4+b4+c4≥(ab)2+(bc)2+(ca)2

(ab)2+(bc)2+(ca)2≥ab2c+a2bc+abc2=abc(a+b+c)

So a4+b4+c4≥abc(a+b+c)

So \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)