Find the maximal positive integer $n$, so that for any real number $x$ we have $\sin^{n}{x}+\cos^{n}{x} \geq \frac{1}{n}$.
1. **Claim**: The maximal \( n \) is \( n = 8 \). We will prove this by showing that \( n \) cannot be greater than 8, and then that \( n \) can be 8.

2. **Proof that \( n \) cannot be greater than 8**:
   - Consider the inequality for \( x = \frac{\pi}{4} \):
     \[
     \sin^n{\frac{\pi}{4}} + \cos^n{\frac{\pi}{4}} = 2\left(\frac{1}{\sqrt{2}}\right)^n = \frac{2}{2^{n/2}} = 2^{1 - \frac{n}{2}}
     \]
   - For the inequality \(\sin^n{x} + \cos^n{x} \geq \frac{1}{n}\) to hold for all \( x \), it must hold for \( x = \frac{\pi}{4} \):
     \[
     2^{1 - \frac{n}{2}} \geq \frac{1}{n}
     \]
   - This simplifies to:
     \[
     n \geq 2^{\frac{n-2}{2}}
     \]
   - The left-hand side grows linearly while the right-hand side grows exponentially. To find the maximum \( n \), we need to check when the inequality fails. For \( n = 9 \):
     \[
     2^{\frac{9-2}{2}} = 2^{3.5} = 8\sqrt{2} \approx 11.3137
     \]
     \[
     9 < 11.3137
     \]
   - This shows that for \( n = 9 \), the inequality \( n \geq 2^{\frac{n-2}{2}} \) does not hold. Therefore, \( n \) cannot be 9 or greater.

3. **Proof that \( n \) can be 8**:
   - We need to show that \(\sin^8{x} + \cos^8{x} \geq \frac{1}{8}\) for all \( x \).
   - Let \( a = \sin^2{x} \) and \( b = \cos^2{x} \). Then \( a + b = 1 \).
   - We need to show \( a^4 + b^4 \geq \frac{1}{8} \).
   - Using the QM-AM inequality:
     \[
     \sqrt{\frac{a^2 + b^2}{2}} \geq \frac{a + b}{2} \implies a^2 + b^2 \geq \frac{1}{2}
     \]
   - Applying QM-AM again:
     \[
     \sqrt{\frac{a^4 + b^4}{2}} \geq \frac{a^2 + b^2}{2} \geq \frac{1}{4} \implies a^4 + b^4 \geq \frac{1}{8}
     \]
   - Therefore, \( \sin^8{x} + \cos^8{x} \geq \frac{1}{8} \).

\(\blacksquare\)

The final answer is \( \boxed{ n = 8 } \).