<s>[INST] 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?
 [/INST] Put $y=0$ giving $a=f(0)$. For non-$0$ $y$,divide the original equation by $y$ [with of course,$a=f(0)$], and let $y$ go to $0$,giving  $f'(x)/2= f'(0)/2+x$. Integrating this,we have $f(x)=xf'(0)+x^2+c$. This satisfies the original equation because $c=a=f(0)$.
</s>