<s>[INST] Minimum of $(xy+a)^2+(bx-cy)^2$ Let $a,b,c>0$. Consider the function
$$f:\mathbb R^2 \to \mathbb R,\quad f(x,y)=(xy+a)^2+(bx-cy)^2.$$
I'm interested in the globel minimum of this function. I could show that for $a \le bc$ the minimum is attained at $(x,y)=(0,0)$ and hence
$$f(0,0)=a^2.$$
For $a>2bc$ however it seems that there exist $x<0<y$ such that $f$ attains the global minimum at $f(x,y)$ and $f(-x,-y)$.
Is this true? How do you prove it and what is $x,y$ in that case?
Of course I computed already the derivative of $f$. But I couldn't solve the equations you get. You get something like
$$
x(y^2+b^2)=y(bc-a),\\
y(x^2+c^2)=x(bc-a).
$$
Edit (one hour later): I could solve it. The minimum is given by $a^2-(a-2bc)$ in case $a \ge 2bc$. I can post an answer if anybody is interested. Edit: That's not quite right. Somewhere is still a mistake. I'll look tomorrow for it.
Edit (next day): I've found my mistake. I forgot a square, the minimum is $a^2-(a-2bc)^2=4bc(a-bc)$.
 [/INST] From completing the square we get
$$
 f(x, y) = (y^2+b^2)x^2 -2y(bc-a) x + (a^2+c^2y^2) \\
= (y^2+b^2) \left[ \left( x - \frac{y(bc-a)}{y^2+b^2}\right)^2 + \frac{a^2+c^2y^2}{y^2+b^2} - \frac{y^2(bc-a)^2}{(y^2+b^2)^2}\right]
$$
and therefore
$$
 f(x, y) \ge a^2+c^2y^2 - \frac{y^2(bc-a)^2}{y^2+b^2}
= \frac{(c y^2 +ab)^2}{y^2+b^2} \, ,
$$
with equality if $x= \frac{y(bc-a)}{y^2+b^2}$. Therefore it suffices to determining the minimum of
$$
 g(u) = \frac{(cu+ab)^2}{u+b^2}
$$
for $u \ge 0$. From
$$
  g'(u) = \frac{(cu+ab)(cu+2b^2c-ab)}{(u+b^2)^2}
$$
we see that:

*

*If $a \le 2bc$ then $g$ is increasing, and the minimum is $g(0) = a^2$.


*If $a > 2bc$ then $g$ has a minimum at $u^*=(a-2bc)b/c$, and the minimum is $g(u^*) = 4bc(a-bc)$.
</s>