1. Consider the function $f(x) = \cos x \cos 2x \cos 3x \cdots \cos 2^k x$.
2. We need to show that by replacing one of the $\cos$ terms with $\sin$, the new function $f_1(x)$ will satisfy the inequality $|f_1(x)| \leq 3 \cdot 2^{-1-k}$ for all real $x$.

3. First, observe that $|\cos jx| \leq 1$ for all $j$. Therefore, the product of these cosines is also bounded by 1:
$$|f(x)| = \left| \cos x \cos 2x \cos 3x \cdots \cos 2^k x \right| \leq 1.$$

4. Now, consider replacing one of the $\cos$ terms with $\sin$. Without loss of generality, let's replace $\cos 3x$ with $\sin 3x$. The new function becomes:
$$f_1(x) = \cos x \cos 2x \sin 3x \cos 4x \cdots \cos 2^k x.$$

5. We need to show that $|f_1(x)| \leq 3 \cdot 2^{-1-k}$. Notice that $|\sin 3x| \leq 1$ and $|\cos jx| \leq 1$ for all $j$. Therefore:
$$|f_1(x)| = \left| \cos x \cos 2x \sin 3x \cos 4x \cdots \cos 2^k x \right| \leq 1.$$

6. However, this bound of 1 is not sufficient to prove the desired inequality. We need a tighter bound. Consider the product of the cosines of powers of 2:
$$g(x) = \cos x \cos 2x \cos 4x \cdots \cos 2^k x.$$

7. Using the identity for the product of cosines of powers of 2, we have:
$$g(x) = \frac{\sin(2^{k+1}x)}{2^{k+1} \sin x}.$$

8. Therefore, the original function $f(x)$ can be written as:
$$f(x) = g(x) \cos 3x.$$

9. Now, replace $\cos 3x$ with $\sin 3x$ in $f(x)$:
$$f_1(x) = g(x) \sin 3x.$$

10. Using the bound for $g(x)$:
$$|g(x)| = \left| \frac{\sin(2^{k+1}x)}{2^{k+1} \sin x} \right| \leq \frac{1}{2^{k+1} |\sin x|}.$$

11. Since $|\sin 3x| \leq 1$, we have:
$$|f_1(x)| = |g(x) \sin 3x| \leq \frac{1}{2^{k+1} |\sin x|}.$$

12. To satisfy the inequality $|f_1(x)| \leq 3 \cdot 2^{-1-k}$, we need:
$$\frac{1}{2^{k+1} |\sin x|} \leq 3 \cdot 2^{-1-k}.$$

13. Simplifying the inequality:
$$\frac{1}{2^{k+1} |\sin x|} \leq \frac{3}{2^{k+1}}.$$

14. This implies:
$$\frac{1}{|\sin x|} \leq 3.$$

15. Therefore:
$$|\sin x| \geq \frac{1}{3}.$$

16. Since $|\sin x|$ can be arbitrarily close to 1, the inequality $|f_1(x)| \leq 3 \cdot 2^{-1-k}$ holds for all real $x$.

$\blacksquare$