Show that the absolute maximum of $f(x, y) = \frac{(ax+by+c)^2}{x^2+y^2+1}$ is $a^2 + b^2 + c^2$

Question

I got this question on my exam today: show that the function $f(x, y) = \dfrac{(ax+by+c)^2}{x^2+y^2+1}$ has an absolute maximum whose value is $a^2 + b^2 + c^2$.

I tried setting the gradient to the zero vector, but it was really hairy. Expanding the numerator also does not seem to help. There also doesn't seem to be any type of coordinate transformation that would help.

The gradient: $$\nabla f(x, y) = -2 \frac{a x + b y + c}{(x^2 + y^2 + 1)^2}\langle x (b y + c) - a (y^2 + 1), y (a x + c) - b (x^2 + 1)\rangle$$

Answer 1

You can notice more or less easily that $f(\frac ac,\frac bc)=a^2+b^2+c^2$.

Since you are hinted that it is the maximum you can set $\begin{cases}x=X+\frac ac\\y=Y+\frac bc\end{cases}$ and calculate (I skip some details...)

$\begin{align}(a^2+b^2+c^2)-f(x,y) &=\dfrac{(b^2+c^2)X^2+(a^2+c^2)Y^2-2abXY}{x^2+y^2+1}\\\\ &=\dfrac{(bX-aY)^2+c^2(X^2+Y^2)}{x^2+y^2+1}\\\\ &\ge 0\end{align}$

The case $c=0$ still need to be handled separately. It is a bit more difficult because in this case the maximum is reached at infinity.

Answer 2

OK... I realize now that expected answer was through the Cauchy-Schwarz inequality. $$\textbf{X} = \langle x, y, 1\rangle $$ $$\textbf{a} = \langle a, b, c\rangle $$ $$\textbf{X}\cdot\textbf{a} \le |\textbf{X}|\cdot|\textbf{a}| $$ And this is equal to: $$ax+by+c \le \sqrt{x^2+y^2+1} \cdot \sqrt{a^2+b^2+c^2}$$ so $$\frac{(ax+by+c)^2}{x^2+y^2+1} \le a^2+b^2+c^2$$