Apply inequality cauchy, we have:

⎧⎩⎨⎪⎪⎪⎪(b+c−a)+(a+c−b)≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√(a+c−b)+(a+b−c)≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√(a+b−c)+(b+c−a)≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪⎪⎪2c≥2(b+c−a)(a+c−b)−−−−−−−−−−−−−−−−−√2a≥2(a+c−b)(a+b−c)−−−−−−−−−−−−−−−−−√2b≥2(a+b−c)(b+c−a)−−−−−−−−−−−−−−−−−√

⇒⎧⎩⎨⎪⎪c2≥(b+c−a)(a+c−b)a2≥(a+c−b)(a+b−c)b2≥(a+b−c)(b+c−a)

⇒(abc)2≥[(b+c−a)(a+c−b)(a+b−c)]2

⇒abc≥(b+c−a)(a+c−b)(a+b−c)

The title must be: a + b = 2

(1st condition): a ≤

 0, b ≥

 0

=> a4≥0;b4≥0

We have: a + b = 2 with a4;b4∈Z

+

=> a4+b4≥a+b≥2

 (proved)

(2nd condition): a ≥

 0; b ≤

 2: Do exactly as the 1st condition (proved)

So ............

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)

a4+b4≥a3b+ab3

<=> a4+b4−a3b−ab3≥0

<=> a3(a−b)−b3(a−b)≥0

<=> (a3−b3)(a−b)≥0

<=> (a−b)2(a2+ab+b2)≥0

Cause ⎧⎩⎨⎪⎪⎪⎪(a−b)2≥0∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0

You have to have a condition : a,b>0

a) (a+b)2≥4ab

<=> a2+2ab+b2≥4ab

<=>  a2−2ab+b2≥0

<=>  (a−b)2≥0

  (It's true)

=> (a+b)2≥4ab

b) Applying Cauchy's inequality , we have

a+b≥2ab−−√

Equation occur <=> a = b

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)

a+b+c = 0

=> (a+b+c)2 = 0

=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0

We have: a2 + b2 + c2 ≥0

⇒2(ab+bc+ca)≤0⇔ab+bc+ca≤0

8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3

⇔8(a3+b3+c3)≥2(a3+b3+c3)+3ab(a+b)+3bc(b+c)+3ca(c+a)

⇔

6(a3+b3+c3)≥3[ab(a+b)+bc(b+c)+ca(c+a)]

⇔

 2a3+2b3+2c3≥ab(a+b)+bc(b+c)+ca(c+a)

We have: a3 + b3 ≥

 a2b + ab2

Prove that the same way as the way of Kaya Renger to prove that: a4+b4≥a3b+ab3

So ⎧⎩⎨⎪⎪a3+b3≥ab(a+b)b3+c3≥bc(b+c)c3+a3≥ca(c+a)

So true. 

=> 8(a3+b3+c3)≥(a+b)3+(b+c)3+(c+a)3

Applying Cauchy's inequality , we have 

(a4+1)+(b4+1)≥2a2+2b2=2(a2+b2)≥2.2ab=4ab

Equation occur <=> a = b = ±1

Apply inequality Cauchy, we have:

a+b≥2ab−−√

⇒{a4+b4≥2a4b4−−−−√c4+d4≥2c4d4−−−−√

⇔a4+b4+c4+d4≥2a2b2+2c2d2=2[(ab)2+(cd)2]

But (ab)2+(cd)2≥2(ab)2⋅(cd)2−−−−−−−−−√=2abcd

⇔a4+b4+c4+d4≥2⋅2abcd=4abcd