How do I prove that the $f(x)$ is positive for all real $x$? $$ \frac {f(x+y) - f(x)}{2}= \frac{f(y)-a}{2} +xy $$ 
for all real $x$ and $y$. If $f(x)$ is differentiable and $f'(0)$ exists for all real permisible values of $a$ and is equal to $\sqrt{5a-1-a^2}$. Prove that $f(x)$ is positive for all real $x$.
I differentiated the equation keeping $x$ constant and then put $y=0$ and then integrated and got $f(x)$ as
$$f(x)= x^2 +x\sqrt{5a-1-a^2}+c$$
by putting $x=y=0$ in 
$$ \frac {f(x+y) - f(x)}{2}= \frac{f(y)-a}{2} +xy $$ 
I got $f(0) =a$ so I finally got the function as 
$$f(x)= x^2 +x\sqrt{5a-1-a^2}+a.$$
Now how should I proceed, will $b^2 -4ac <0 $ help?

Are you sure about your expression for $f(x)$? (doing it very quickly, it looks like the coefficient of $x^2$ should be one, not $\frac{1}{2}$, and similarly: no $2$ coefficient in front of the square root).
Assuming this, then it follows that the quadratic polynomial
$$
f(x) = x^2 + x\sqrt{5a-1-a^2} + a
$$
is always positive. Indeed, it has no roots: $\Delta = - (a^2-a+1) < 0$, and therefore always has (strictly) same sign as the leading coefficient (of $x^2$), which is $1$.
