What is the least real number $C$ that satisfies $\sin x \cos x \leq C(\sin^6x+\cos^6x)$ for every real number $x$?

$ \textbf{(A)}\ \sqrt3 \qquad \textbf{(B)}\ 2\sqrt2 \qquad \textbf{(C)}\ \sqrt 2 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \text{None}$
1. We start with the given inequality:
   \[
   \sin x \cos x \leq C(\sin^6 x + \cos^6 x)
   \]

2. Using the double-angle identity for sine, we rewrite \(\sin x \cos x\) as:
   \[
   \sin x \cos x = \frac{1}{2} \sin 2x
   \]
   Thus, the inequality becomes:
   \[
   \frac{1}{2} \sin 2x \leq C(\sin^6 x + \cos^6 x)
   \]

3. Next, we use the identity \(\sin^2 x + \cos^2 x = 1\) to express \(\sin^6 x + \cos^6 x\) in a different form. We use the identity for the sum of cubes:
   \[
   a^3 + b^3 = (a + b)(a^2 - ab + b^2)
   \]
   where \(a = \sin^2 x\) and \(b = \cos^2 x\). Therefore:
   \[
   \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2)
   \]
   Since \(\sin^2 x + \cos^2 x = 1\), we have:
   \[
   \sin^6 x + \cos^6 x = 1 \cdot (1 - 3\sin^2 x \cos^2 x) = 1 - 3\sin^2 x \cos^2 x
   \]

4. Substituting this back into the inequality, we get:
   \[
   \frac{1}{2} \sin 2x \leq C(1 - 3\sin^2 x \cos^2 x)
   \]
   Using the double-angle identity again, \(\sin^2 x \cos^2 x = \left(\frac{1}{2} \sin 2x\right)^2 = \frac{1}{4} \sin^2 2x\), we substitute:
   \[
   \frac{1}{2} \sin 2x \leq C \left(1 - 3 \cdot \frac{1}{4} \sin^2 2x\right)
   \]
   Simplifying inside the parentheses:
   \[
   \frac{1}{2} \sin 2x \leq C \left(1 - \frac{3}{4} \sin^2 2x\right)
   \]

5. To find the least \(C\), we need to consider the maximum value of \(\sin 2x\), which is 1. Let \(\sin 2x = t\), where \(t \in [-1, 1]\). The inequality becomes:
   \[
   \frac{1}{2} t \leq C \left(1 - \frac{3}{4} t^2\right)
   \]

6. Rearranging the inequality:
   \[
   \frac{1}{2} t \leq C - \frac{3}{4} Ct^2
   \]
   \[
   \frac{1}{2} t + \frac{3}{4} Ct^2 \leq C
   \]

7. To find the least \(C\), we need to solve for \(C\) when \(t = 1\) (since \(\sin 2x\) can reach its maximum value of 1):
   \[
   \frac{1}{2} \cdot 1 + \frac{3}{4} C \cdot 1^2 \leq C
   \]
   \[
   \frac{1}{2} + \frac{3}{4} C \leq C
   \]
   \[
   \frac{1}{2} \leq C - \frac{3}{4} C
   \]
   \[
   \frac{1}{2} \leq \frac{1}{4} C
   \]
   \[
   2 \leq C
   \]

Thus, the least value of \(C\) that satisfies the inequality for all \(x\) is \(C = 2\).

The final answer is \(\boxed{2}\).