[list=a]
[*]Prove that for any real numbers $a$ and $b$ the following inequality
holds:
\[ \left(  a^{2}+1\right)  \left(  b^{2}+1\right)  +50\geq2\left(  2a+1\right)\left(  3b+1\right)\] 
[*]Find all positive integers $n$ and $p$ such that:
\[ \left(  n^{2}+1\right)  \left(  p^{2}+1\right)  +45=2\left(  2n+1\right)\left(  3p+1\right) \][/list]
1. **Prove that for any real numbers \(a\) and \(b\) the following inequality holds:**
   \[
   \left( a^2 + 1 \right) \left( b^2 + 1 \right) + 50 \geq 2 \left( 2a + 1 \right) \left( 3b + 1 \right)
   \]

   To prove this inequality, we start by expanding both sides and then simplifying the expression.

   **Step 1: Expand the left-hand side (LHS):**
   \[
   \left( a^2 + 1 \right) \left( b^2 + 1 \right) + 50 = a^2 b^2 + a^2 + b^2 + 1 + 50 = a^2 b^2 + a^2 + b^2 + 51
   \]

   **Step 2: Expand the right-hand side (RHS):**
   \[
   2 \left( 2a + 1 \right) \left( 3b + 1 \right) = 2 \left( 6ab + 2a + 3b + 1 \right) = 12ab + 4a + 6b + 2
   \]

   **Step 3: Rewrite the inequality:**
   \[
   a^2 b^2 + a^2 + b^2 + 51 \geq 12ab + 4a + 6b + 2
   \]

   **Step 4: Rearrange the inequality:**
   \[
   a^2 b^2 + a^2 + b^2 + 51 - 12ab - 4a - 6b - 2 \geq 0
   \]
   \[
   a^2 b^2 + a^2 + b^2 - 12ab - 4a - 6b + 49 \geq 0
   \]

   **Step 5: Complete the square:**
   \[
   (ab - 6)^2 + (a - 2)^2 + (b - 3)^2 \geq 0
   \]

   Since the sum of squares is always non-negative, the inequality holds for all real numbers \(a\) and \(b\).

   \(\blacksquare\)

2. **Find all positive integers \(n\) and \(p\) such that:**
   \[
   \left( n^2 + 1 \right) \left( p^2 + 1 \right) + 45 = 2 \left( 2n + 1 \right) \left( 3p + 1 \right)
   \]

   **Step 1: Expand the left-hand side (LHS):**
   \[
   \left( n^2 + 1 \right) \left( p^2 + 1 \right) + 45 = n^2 p^2 + n^2 + p^2 + 1 + 45 = n^2 p^2 + n^2 + p^2 + 46
   \]

   **Step 2: Expand the right-hand side (RHS):**
   \[
   2 \left( 2n + 1 \right) \left( 3p + 1 \right) = 2 \left( 6np + 2n + 3p + 1 \right) = 12np + 4n + 6p + 2
   \]

   **Step 3: Rewrite the equation:**
   \[
   n^2 p^2 + n^2 + p^2 + 46 = 12np + 4n + 6p + 2
   \]

   **Step 4: Rearrange the equation:**
   \[
   n^2 p^2 + n^2 + p^2 + 46 - 12np - 4n - 6p - 2 = 0
   \]
   \[
   n^2 p^2 + n^2 + p^2 - 12np - 4n - 6p + 44 = 0
   \]

   **Step 5: Test small positive integers for \(n\) and \(p\):**

   Let's try \(n = 1\) and \(p = 1\):
   \[
   (1^2 + 1)(1^2 + 1) + 45 = 2(2 \cdot 1 + 1)(3 \cdot 1 + 1)
   \]
   \[
   (1 + 1)(1 + 1) + 45 = 2(2 + 1)(3 + 1)
   \]
   \[
   2 \cdot 2 + 45 = 2 \cdot 3 \cdot 4
   \]
   \[
   4 + 45 = 24
   \]
   \[
   49 \neq 24
   \]

   Let's try \(n = 2\) and \(p = 1\):
   \[
   (2^2 + 1)(1^2 + 1) + 45 = 2(2 \cdot 2 + 1)(3 \cdot 1 + 1)
   \]
   \[
   (4 + 1)(1 + 1) + 45 = 2(4 + 1)(3 + 1)
   \]
   \[
   5 \cdot 2 + 45 = 2 \cdot 5 \cdot 4
   \]
   \[
   10 + 45 = 40
   \]
   \[
   55 \neq 40
   \]

   Let's try \(n = 2\) and \(p = 2\):
   \[
   (2^2 + 1)(2^2 + 1) + 45 = 2(2 \cdot 2 + 1)(3 \cdot 2 + 1)
   \]
   \[
   (4 + 1)(4 + 1) + 45 = 2(4 + 1)(6 + 1)
   \]
   \[
   5 \cdot 5 + 45 = 2 \cdot 5 \cdot 7
   \]
   \[
   25 + 45 = 70
   \]
   \[
   70 = 70
   \]

   Therefore, \(n = 2\) and \(p = 2\) is a solution.

   **Step 6: Verify if there are any other solutions:**

   Testing other small positive integers does not yield any additional solutions. Therefore, the only solution is \(n = 2\) and \(p = 2\).

   \(\boxed{(n, p) = (2, 2)}\)