a) If $K$ is a finite extension of the field $F$ and $K=F(\alpha,\beta)$ show that $[K: F]\leq [F(\alpha): F][F(\beta): F]$

b) If $gcd([F(\alpha): F],[F(\beta): F])=1$ then does the above inequality always become equality?

c) By giving an example show that if $gcd([F(\alpha): F],[F(\beta): F])\neq 1$ then equality might happen.
### Part (a)
To show that if \( K \) is a finite extension of the field \( F \) and \( K = F(\alpha, \beta) \), then \([K: F] \leq [F(\alpha): F][F(\beta): F]\):

1. Let \( [F(\alpha): F] = m \) and \( [F(\beta): F] = n \). This means that the minimal polynomials of \(\alpha\) and \(\beta\) over \(F\) have degrees \(m\) and \(n\) respectively.

2. Consider the field extensions \( F \subseteq F(\alpha) \subseteq F(\alpha, \beta) = K \) and \( F \subseteq F(\beta) \subseteq F(\alpha, \beta) = K \).

3. By the Tower Law (or the multiplicativity of degrees of field extensions), we have:
   \[
   [K: F] = [K: F(\alpha)][F(\alpha): F]
   \]
   and
   \[
   [K: F] = [K: F(\beta)][F(\beta): F]
   \]

4. Since \( F(\alpha) \subseteq K \) and \( F(\beta) \subseteq K \), we have:
   \[
   [K: F(\alpha)] \leq [F(\alpha, \beta): F(\alpha)]
   \]
   and
   \[
   [K: F(\beta)] \leq [F(\alpha, \beta): F(\beta)]
   \]

5. Therefore, combining these inequalities, we get:
   \[
   [K: F] \leq [F(\alpha, \beta): F(\alpha)][F(\alpha): F]
   \]
   and
   \[
   [K: F] \leq [F(\alpha, \beta): F(\beta)][F(\beta): F]
   \]

6. Since \( [F(\alpha, \beta): F(\alpha)] \leq [F(\beta): F] \) and \( [F(\alpha, \beta): F(\beta)] \leq [F(\alpha): F] \), we have:
   \[
   [K: F] \leq [F(\alpha): F][F(\beta): F]
   \]

Thus, we have shown that \([K: F] \leq [F(\alpha): F][F(\beta): F]\).

### Part (b)
If \( \gcd([F(\alpha): F], [F(\beta): F]) = 1 \), then does the above inequality always become equality?

1. Let \( [F(\alpha): F] = m \) and \( [F(\beta): F] = n \) with \(\gcd(m, n) = 1\).

2. Since \( \gcd(m, n) = 1 \), the degrees \( m \) and \( n \) are coprime.

3. By the Chinese Remainder Theorem, the field \( F(\alpha, \beta) \) must have a degree over \( F \) that is the product of the degrees of the individual extensions, i.e., \( mn \).

4. Therefore, we have:
   \[
   [K: F] = [F(\alpha, \beta): F] = mn
   \]

Thus, the inequality becomes equality when \(\gcd([F(\alpha): F], [F(\beta): F]) = 1\).

### Part (c)
To give an example where \(\gcd([F(\alpha): F], [F(\beta): F]) \neq 1\) and equality might still happen:

1. Consider \( F = \mathbb{Q} \), \( \alpha = \sqrt{2} \), and \( \beta = \sqrt{3} \).

2. Here, \( [\mathbb{Q}(\sqrt{2}): \mathbb{Q}] = 2 \) and \( [\mathbb{Q}(\sqrt{3}): \mathbb{Q}] = 2 \).

3. The degrees are not coprime since \(\gcd(2, 2) = 2\).

4. However, \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3}) \) is a degree 4 extension over \(\mathbb{Q}\).

5. Therefore, we have:
   \[
   [\mathbb{Q}(\sqrt{2}, \sqrt{3}): \mathbb{Q}] = 4 = 2 \times 2
   \]

Thus, even though \(\gcd([F(\alpha): F], [F(\beta): F]) \neq 1\), the equality \([K: F] = [F(\alpha): F][F(\beta): F]\) still holds in this case.

The final answer is \( \boxed{ [K: F] \leq [F(\alpha): F][F(\beta): F] } \) and equality holds if \(\gcd([F(\alpha): F], [F(\beta): F]) = 1\).