Find all real numbers $k$ for which the inequality 

$(1+t)^k (1-t)^{1-k} \leq 1$ 

is true for every real number $t \in (-1, 1)$.
1. **Verification for \( k = \frac{1}{2} \):**
   We need to show that for \( k = \frac{1}{2} \), the inequality holds for all \( t \in (-1, 1) \). 

   \[
   (1+t)^{\frac{1}{2}} (1-t)^{\frac{1}{2}} = \sqrt{(1+t)(1-t)} = \sqrt{1-t^2}
   \]

   Since \( t \in (-1, 1) \), we have \( 1 - t^2 \in (0, 1] \). Therefore,

   \[
   \sqrt{1-t^2} \leq 1
   \]

   This shows that the inequality holds for \( k = \frac{1}{2} \).

2. **Uniqueness of \( k = \frac{1}{2} \):**
   Suppose \( k \neq \frac{1}{2} \). We will show that there exists some \( t \in (-1, 1) \) such that the inequality does not hold.

   Consider the expression \( (1+t)^k (1-t)^{1-k} \). For small \( t \), we can use the first-order Taylor expansion around \( t = 0 \):

   \[
   (1+t)^k \approx 1 + kt + \frac{k(k-1)}{2}t^2
   \]
   \[
   (1-t)^{1-k} \approx 1 - (1-k)t + \frac{(1-k)(-k)}{2}t^2
   \]

   Multiplying these approximations:

   \[
   (1+t)^k (1-t)^{1-k} \approx \left(1 + kt + \frac{k(k-1)}{2}t^2\right) \left(1 - (1-k)t + \frac{(1-k)(-k)}{2}t^2\right)
   \]

   Expanding and keeping terms up to the second order:

   \[
   \approx 1 + kt - (1-k)t + \frac{k(k-1)}{2}t^2 + \frac{(1-k)(-k)}{2}t^2 + \text{higher order terms}
   \]
   \[
   = 1 + (k - (1-k))t + \left(\frac{k(k-1)}{2} + \frac{(1-k)(-k)}{2}\right)t^2
   \]
   \[
   = 1 + (2k-1)t + \left(\frac{k^2 - k + k - k^2}{2}\right)t^2
   \]
   \[
   = 1 + (2k-1)t
   \]

   For \( k \neq \frac{1}{2} \), the term \( (2k-1)t \) is non-zero. Therefore, for sufficiently small \( t \), the expression \( 1 + (2k-1)t \) can be greater than 1 or less than 1, depending on the sign of \( 2k-1 \). This means the inequality \( (1+t)^k (1-t)^{1-k} \leq 1 \) does not hold for all \( t \in (-1, 1) \).

   Hence, \( k = \frac{1}{2} \) is the only value that satisfies the inequality for all \( t \in (-1, 1) \).

\(\blacksquare\)

The final answer is \( \boxed{ k = \frac{1}{2} } \).