For which real numbers $c$ is
$$\frac{e^x +e^{-x} }{2} \leq e^{c x^2 }$$
for all real $x?$
1. We start with the given inequality:
   \[
   \frac{e^x + e^{-x}}{2} \leq e^{c x^2}
   \]
   for all real \( x \).

2. Recognize that the left-hand side, \(\frac{e^x + e^{-x}}{2}\), is the hyperbolic cosine function, \(\cosh(x)\):
   \[
   \cosh(x) = \frac{e^x + e^{-x}}{2}
   \]

3. The Taylor series expansion for \(\cosh(x)\) is:
   \[
   \cosh(x) = \sum_{n=0}^\infty \frac{x^{2n}}{(2n)!}
   \]

4. We need to compare this with the right-hand side, \(e^{c x^2}\). The Taylor series expansion for \(e^{c x^2}\) is:
   \[
   e^{c x^2} = \sum_{n=0}^\infty \frac{(c x^2)^n}{n!} = \sum_{n=0}^\infty \frac{c^n x^{2n}}{n!}
   \]

5. To find the condition on \(c\), we compare the coefficients of \(x^{2n}\) in both series. For the inequality to hold for all \(x\), each term in the series expansion of \(\cosh(x)\) must be less than or equal to the corresponding term in the series expansion of \(e^{c x^2}\).

6. Consider the first few terms of the series:
   \[
   \cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots
   \]
   \[
   e^{c x^2} = 1 + c x^2 + \frac{(c x^2)^2}{2!} + \cdots
   \]

7. For the inequality to hold, we need:
   \[
   \frac{x^2}{2!} \leq c x^2
   \]
   Dividing both sides by \(x^2\) (assuming \(x \neq 0\)):
   \[
   \frac{1}{2} \leq c
   \]
   Thus, \(c \geq \frac{1}{2}\).

8. To confirm, we check the higher-order terms. For \(x^4\):
   \[
   \frac{x^4}{4!} \leq \frac{(c x^2)^2}{2!} = \frac{c^2 x^4}{2}
   \]
   Dividing both sides by \(x^4\):
   \[
   \frac{1}{24} \leq \frac{c^2}{2}
   \]
   \[
   \frac{1}{24} \leq \frac{c^2}{2} \implies \frac{1}{12} \leq c^2 \implies c \geq \frac{1}{2}
   \]
   This confirms that \(c \geq \frac{1}{2}\) is sufficient for the inequality to hold for all \(x\).

9. Therefore, the inequality \(\frac{e^x + e^{-x}}{2} \leq e^{c x^2}\) holds if and only if \(c \geq \frac{1}{2}\).

The final answer is \( \boxed{ c \geq \frac{1}{2} } \)