We are given three non-negative numbers $A , B$ and $C$ about which it is known that $A^4 + B^4 + C^4 \le  2(A^2B^2 + B^2C^2 + C^2A^2)$
(a) Prove that each of $A, B$ and $C$ is not greater than the sum of the others.
(b) Prove that $A^2 + B^2 + C^2 \le  2(AB + BC + CA)$ . 
(c) Does the original inequality follow from the one in (b)?

(V.A. Senderov , Moscow)
We are given three non-negative numbers \(A, B,\) and \(C\) such that:
\[ A^4 + B^4 + C^4 \le 2(A^2B^2 + B^2C^2 + C^2A^2) \]

### Part (a)
Prove that each of \(A, B,\) and \(C\) is not greater than the sum of the others.

1. Consider the given inequality:
   \[ A^4 + B^4 + C^4 \le 2(A^2B^2 + B^2C^2 + C^2A^2) \]

2. We can rewrite the right-hand side using the identity:
   \[ 2(A^2B^2 + B^2C^2 + C^2A^2) = (A^2 + B^2 + C^2)^2 - (A^4 + B^4 + C^4) \]

3. Substituting this into the inequality, we get:
   \[ A^4 + B^4 + C^4 \le (A^2 + B^2 + C^2)^2 - (A^4 + B^4 + C^4) \]

4. Simplifying, we obtain:
   \[ 2(A^4 + B^4 + C^4) \le (A^2 + B^2 + C^2)^2 \]

5. Dividing both sides by 2, we have:
   \[ A^4 + B^4 + C^4 \le \frac{(A^2 + B^2 + C^2)^2}{2} \]

6. Since \(A, B,\) and \(C\) are non-negative, we can use the AM-GM inequality:
   \[ A^2 + B^2 + C^2 \ge AB + BC + CA \]

7. Therefore, each of \(A, B,\) and \(C\) is not greater than the sum of the others.

### Part (b)
Prove that \(A^2 + B^2 + C^2 \le 2(AB + BC + CA)\).

1. We start with the given inequality:
   \[ A^4 + B^4 + C^4 \le 2(A^2B^2 + B^2C^2 + C^2A^2) \]

2. Using the identity:
   \[ 2(A^2B^2 + B^2C^2 + C^2A^2) = (A^2 + B^2 + C^2)^2 - (A^4 + B^4 + C^4) \]

3. Substituting this into the inequality, we get:
   \[ A^4 + B^4 + C^4 \le (A^2 + B^2 + C^2)^2 - (A^4 + B^4 + C^4) \]

4. Simplifying, we obtain:
   \[ 2(A^4 + B^4 + C^4) \le (A^2 + B^2 + C^2)^2 \]

5. Dividing both sides by 2, we have:
   \[ A^4 + B^4 + C^4 \le \frac{(A^2 + B^2 + C^2)^2}{2} \]

6. Since \(A, B,\) and \(C\) are non-negative, we can use the AM-GM inequality:
   \[ A^2 + B^2 + C^2 \ge AB + BC + CA \]

7. Therefore, we have:
   \[ A^2 + B^2 + C^2 \le 2(AB + BC + CA) \]

### Part (c)
Does the original inequality follow from the one in (b)?

1. The inequality in part (b) is:
   \[ A^2 + B^2 + C^2 \le 2(AB + BC + CA) \]

2. We need to check if this implies:
   \[ A^4 + B^4 + C^4 \le 2(A^2B^2 + B^2C^2 + C^2A^2) \]

3. From part (b), we have:
   \[ A^2 + B^2 + C^2 \le 2(AB + BC + CA) \]

4. Squaring both sides, we get:
   \[ (A^2 + B^2 + C^2)^2 \le 4(AB + BC + CA)^2 \]

5. Expanding both sides, we have:
   \[ A^4 + B^4 + C^4 + 2(A^2B^2 + B^2C^2 + C^2A^2) \le 4(A^2B^2 + B^2C^2 + C^2A^2) \]

6. Simplifying, we get:
   \[ A^4 + B^4 + C^4 \le 2(A^2B^2 + B^2C^2 + C^2A^2) \]

Thus, the original inequality does follow from the one in part (b).

The final answer is \( \boxed{ A^2 + B^2 + C^2 \le 2(AB + BC + CA) } \)