Find the maximum value of the real constant $C$ such that $x^{2}+y^{2}+1\geq C(x+y)$, and $ x^{2}+y^{2}+xy+1\geq  C(x+y)$ for all reals $x,y$.
To find the maximum value of the real constant \( C \) such that the inequalities \( x^2 + y^2 + 1 \geq C(x + y) \) and \( x^2 + y^2 + xy + 1 \geq C(x + y) \) hold for all real numbers \( x \) and \( y \), we will analyze each inequality separately and then find the common maximum value of \( C \).

### Step 1: Analyzing \( x^2 + y^2 + 1 \geq C(x + y) \)

1. Consider the inequality \( x^2 + y^2 + 1 \geq C(x + y) \).
2. To find the maximum \( C \), we can test specific values of \( x \) and \( y \). Let \( x = y = \frac{\sqrt{2}}{2} \).
3. Substituting \( x = y = \frac{\sqrt{2}}{2} \) into the inequality:
   \[
   \left(\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + 1 \geq C \left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right)
   \]
4. Simplifying the left-hand side:
   \[
   \frac{1}{2} + \frac{1}{2} + 1 = 2
   \]
5. Simplifying the right-hand side:
   \[
   C \cdot \sqrt{2}
   \]
6. Therefore, the inequality becomes:
   \[
   2 \geq C \cdot \sqrt{2}
   \]
7. Solving for \( C \):
   \[
   C \leq \frac{2}{\sqrt{2}} = \sqrt{2}
   \]

### Step 2: Analyzing \( x^2 + y^2 + xy + 1 \geq C(x + y) \)

1. Consider the inequality \( x^2 + y^2 + xy + 1 \geq C(x + y) \).
2. To find the maximum \( C \), we can test specific values of \( x \) and \( y \). Let \( x = y = \frac{\sqrt{3}}{3} \).
3. Substituting \( x = y = \frac{\sqrt{3}}{3} \) into the inequality:
   \[
   \left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)^2 + \left(\frac{\sqrt{3}}{3}\right)\left(\frac{\sqrt{3}}{3}\right) + 1 \geq C \left( \frac{2\sqrt{3}}{3} \right)
   \]
4. Simplifying the left-hand side:
   \[
   \frac{1}{3} + \frac{1}{3} + \frac{1}{3} + 1 = 2
   \]
5. Simplifying the right-hand side:
   \[
   C \cdot \frac{2\sqrt{3}}{3}
   \]
6. Therefore, the inequality becomes:
   \[
   2 \geq C \cdot \frac{2\sqrt{3}}{3}
   \]
7. Solving for \( C \):
   \[
   C \leq \frac{3}{\sqrt{3}} = \sqrt{3}
   \]

### Conclusion

The maximum value of \( C \) must satisfy both inequalities. Therefore, the maximum value of \( C \) is the minimum of the two values found:
\[
C \leq \min(\sqrt{2}, \sqrt{3}) = \sqrt{2}
\]

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