How do I prove that the $f(x)$ is positive for all real $x$?

Question

$$ \frac {f(x+y) - f(x)}{2}= \frac{f(y)-a}{2} +xy $$ 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?

Answer 1

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 $1$.

Answer 2

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 $c=a=f(0)$.