If x is real, then the maximum value of $y=2(a-x)\left(x+\sqrt{x^2+b^2}\right)$ is

Let $t=x+\sqrt{x^2+b^2}$

$\Rightarrow \frac{1}{t}=\frac{1}{x+\sqrt{x^2+b^2}}=\frac{\sqrt{x^2+b^2}-x}{b^2}$

$\therefore t-\frac{b^2}{t}=2x$ and $t+\frac{b^2}{t}=2\sqrt{x^2+b^2}$

Thus $2(a-x)\left(x+\sqrt{x^2+b^2}\right)=\left(2a-t+\frac{b^2}{t}\right)\cdot \left(t\right)$

$\begin{array}{l}=2at-t^2+b^2=a^2+b^2-\left(a^2-2at+t^2\right)\\ =a^2+b^2-(a-t{)}^2\end{array}$

Therefore, $y=2(a-x)\left(x+\sqrt{x^2+b^2}\right)\le a^2+b^2$. Hence, the maximum value of y is $a^2+b^2$. This value is attained at $t=a$ or $x=\left(a^2-b^2\right)/2a$