Answer     $36$

Solving $2x=\frac{x^2+4}{2}$ leads to $x=2$. Therefore, we have $f(2) = 4$.

Let $f(x)=ax^2 + bx + c$. Setting $x=-2, 2$ respectively gives $4a - 2b + c= 0$ and $4a+2b+c=4$. Substracting the second equation from the first one leads to $b=1$. Hence $f(x)=ax^2 + x + c$ and $4a + c = 2$.

Furthermore, replacing $c$ with $2-4a$ gives $f(x) = ax^2 + x + 2-4a$. Because $2x \le f(x)$ always holds, we have

$$ax^2 + x + 2 - 4a \ge 2x \implies ax^2 - x - 2 - 4a \gt 0 $$

Therefore, we must have $a > 0$ and $\Delta = 1^2 - 4 a(2-4a)\lt 0 \implies (4a-1)^2 \lt 0 \implies a=\frac{1}{4}$. 

It follows that $c=1$ and $f(x)=\frac{1}{4}x^2 + x + 1$. Heance $f(10)=\boxed{36}$.