Find all $ f:N\rightarrow N$, such that $\forall m,n\in N $ 
$ 2f(mn) \geq f(m^2+n^2)-f(m)^2-f(n)^2 \geq 2f(m)f(n) $
1. Let \( P(m, n) \) be the assertion of the given functional equation:
   \[
   2f(mn) \geq f(m^2 + n^2) - f(m)^2 - f(n)^2 \geq 2f(m)f(n)
   \]

2. Consider \( P(1, 1) \):
   \[
   2f(1) \geq f(2) - 2f(1)^2 \geq 2f(1)^2
   \]
   This simplifies to:
   \[
   2f(1) \geq f(2) - 2f(1)^2 \quad \text{and} \quad f(2) - 2f(1)^2 \geq 2f(1)^2
   \]
   Combining these, we get:
   \[
   2f(1) \geq f(2) - 2f(1)^2 \geq 2f(1)^2
   \]
   From \( 2f(1) \geq 2f(1)^2 \), we have:
   \[
   f(1) \geq f(1)^2
   \]
   Since \( f(1) \in \mathbb{N} \), the only solution is \( f(1) = 1 \).

3. Consider \( P(m, 1) \):
   \[
   2f(m) + f(m)^2 + f(1) \geq f(m^2 + 1) \geq 2f(m) + f(m)^2 + 1
   \]
   Simplifying, we get:
   \[
   (f(m) + 1)^2 \geq f(m^2 + 1) \geq (f(m) + 1)^2
   \]
   Therefore:
   \[
   f(m^2 + 1) = (f(m) + 1)^2
   \]

4. Take \( m = 1 \) to get:
   \[
   f(2) = (f(1) + 1)^2 = 4
   \]

5. Now, it is easy to get:
   \[
   f(5) = (f(2) + 1)^2 = 25
   \]
   \[
   f(26) = (f(5) + 1)^2 = 676
   \]

6. From the given inequality:
   \[
   2f(mn) \geq f(m^2 + n^2) - f(m)^2 - f(n)^2 \geq 2f(m)f(n)
   \]
   We get:
   \[
   f(mn) \geq f(m)f(n) \quad \text{(denoted as } \spadesuit \text{)}
   \]

7. Consider \( f(2m^2) \geq f(2)f(m^2) \):
   \[
   f(2m^2) \geq 4f(m^2)
   \]

8. Consider \( P(m, m) \):
   \[
   2f(m^2) \geq f(2m^2) - 2f(m)^2
   \]
   Multiplying this by 2, we get:
   \[
   4f(m^2) \geq 2f(2m^2) - 4f(m)^2
   \]
   We have already proved \( f(2m^2) \geq 4f(m^2) \). Therefore:
   \[
   f(2m^2) \geq 2f(2m^2) - 4f(m)^2 \implies 4f(m)^2 \geq f(2m^2)
   \]

9. From \( \spadesuit \), we have:
   \[
   f(2m^2) \geq f(2)f(m^2) = 4f(m)^2
   \]
   Hence:
   \[
   f(m)^2 \geq f(m^2)
   \]
   But \( \spadesuit \) implies:
   \[
   f(m^2) \geq f(m)^2
   \]
   Therefore, we have:
   \[
   f(m^2) = f(m)^2
   \]

10. Considering \( P(m, m) \), it is easy to see that:
    \[
    f(2m^2) = 4f(m)^2
    \]

11. Now, it is easy to see that:
    \[
    f(2^k) = 2^{2k} \quad \forall k \in \mathbb{N}
    \]

12. In \( \spadesuit \), consider \( P(m, 2m) \):
    \[
    f(2m^2) = f(m)f(2m)
    \]
    But:
    \[
    f(2m^2) = 4f(m)^2
    \]
    Therefore:
    \[
    4f(m)^2 = f(m)f(2m) \implies f(2m) = 4f(m) \quad \forall m \in \mathbb{N}
    \]

13. We have proved:
    \[
    (f(m) + 1)^2 = f(m^2 + 1)
    \]
    Take \( m = 3 \) to get:
    \[
    (f(3) + 1)^2 = f(10)
    \]
    We have proved \( f(5) = 25 \) and \( f(2n) = 4f(n) \), therefore:
    \[
    f(10) = 4f(5) = 100 \implies f(10) = 100
    \]
    Therefore:
    \[
    (f(3) + 1)^2 = 100 \implies f(3) = 9
    \]

14. We have proved:
    \[
    f(1) = 1, \quad f(2) = 4, \quad f(3) = 9, \quad f(4) = 16, \quad f(5) = 25
    \]
    We now claim:
    \[
    f(n) = n^2 \quad \forall n \in \mathbb{N}
    \]

15. Let's say it is true till some positive integers \( n \). We have already proved \( f(2n) = 4f(n) \), hence, if we know \( f(n) \), we can find \( f(2n) \).

16. Let \( m = 2x + 1 \) be an odd integer. Then:
    \[
    (f(m) + 1)^2 = f(m^2 + 1) \implies (f(m) + 1)^2 = f(4x^2 + 4x + 2) = 2f(2x^2 + 2x + 1)
    \]
    \[
    = 2f(x^2 + (x + 1)^2)
    \]

17. Taking \( P(x, x + 1) \) in:
    \[
    2f(mn) \geq f(m^2 + n^2) - f(m)^2 - f(n)^2 \geq 2f(m)f(n)
    \]
    We can find \( f(x^2 + (x + 1)^2) \).

18. We have proved that \( f(m^2 + 1) = (f(m) + 1)^2 \). It is easy to show that for all the numbers in the sequence, \( f(u_n) = u_n^2 \).

19. Taking \( P(m, u_n) \), we can conclude:
    \[
    f(m^2 + u_n^2) = (f(m) + u_n^2)^2 \quad \forall n \in \mathbb{N}
    \]
    (All the numbers in this sequence).

20. To complete the induction, we only have to prove the "odd integers case". Let \( m = 2x + 1 \) be an odd integer. Taking \( n = 1 \), we get \( u_n = 1 \). If we take \( n = 1 \), it only remains to find the value of \( f(x(x + 1)) \).

21. Notice that taking even \( n \) does not help, so we will take only odd \( n \)'s. Taking \( n = 3 \), we get \( u_n = 5 \). We have proved:
    \[
    f(m^2 + u_n^2) = (f(m) + u_n^2)^2 \quad \forall n \in \mathbb{N}
    \]
    Therefore:
    \[
    f(m^2 + 25) = (f(m) + 25)^2
    \]
    Take \( m = 2x + 1 \):
    \[
    f(4x^2 + 4x + 26) = (f(2x + 1) + 25)^2 = 2f(2x^2 + 2x + 13) = 2f((x + 3)^2 + (x - 2)^2)
    \]

22. Take \( P(x + 3, x - 2) \) in the original inequality, it remains to find the value of \( f(x^2 + x - 6) \).

23. Induction Complete. Therefore:
    \[
    f(n) = n^2 \quad \forall n \in \mathbb{N}
    \]

The final answer is \( \boxed{ f(n) = n^2 \quad \forall n \in \mathbb{N} } \).