Minimum of $(xy+a)^2+(bx-cy)^2$

Question

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 $a^2-(a-2bc)^2=4bc(a-bc)$.

Answer 1

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 $g(u^*) = 4bc(a-bc)$.

Answer 2

We have

• $f_x=2xy^2+2b^2x+2(a-bc)y=0$
• $f_y=2x^2y+2c^2y+2(a-bc)x=0$

one critical point is given by $(x,y)=(0,0)$ and for $xy\neq 0$ we obtain

• $xf_x=2x^2y^2+2b^2x^2+2(a-bc)xy=0$
• $yf_y=2x^2y^2+2c^2y^2+2(a-bc)xy=0$

which by $xf_x-yf_y$ leads to $x=\pm \frac c b y$ and then

$$f_x\left(\pm \frac c b y,y\right)=\pm 2\frac c b y^3\pm 2 bc y+2(a-bc)y=0$$

$$y^2+b^2\pm \frac{ab}c\mp b^2 =0 \implies y_0=\pm \sqrt{\frac{b(a-2bc)}{c}}\quad x_0=\mp\sqrt{\frac{c(a-2bc)}{b}}$$

Therefore for $a\le2bc$ we only have $(x,y)=(0,0)$ as a critical point which by inspection is a global minimum for the function.

For $a>2bc$ let consider

• $f_{xx}=2y^2+2b^2$
• $f_{yy}=2x^2+2c^2$
• $f_{xy}=4xy+2(a-bc)$

from which we find that the Hessian determinant is equal to

$$\det(H(x_0,y_0)) = f_{xx}(x_0,y_0)f_{yy}(x_0,y_0) - \left( f_{xy}(x_0,y_0) \right)^2 =128bc(a-2bc)>0$$

which indicates that $(x_0,y_0)$ are local minimum points with $f(x_0,y_0)=4bc(a-bc)$.

Answer 3

For completeness I post my solution even though there is already one by user.

Wanting the first derivative to vanish you get $$ (xy+a)y=-b(bx-cy),\\ (xy+a)x=c(bx-cy). $$ So this gives you already $$\frac{c}{x}=-\frac{b}{y}$$ or equivalently $$y=-\frac{b}{c}x.$$ Here I want to mention that in user's approach you couldn't tell the sign (but we can).

Now we have $$f(x,-\frac{b}{c}x) = \frac{b^2}{c^2}z^2+(4b^2-2\frac{ab}{c})z+a^2$$ where we write $z:=x^2$. In $z$ this is a parabola whose minimum is attained at $$z_0:= \frac{-(4b^2-2\frac{ab}{c})}{2\frac{b^2}{c^2}}=\frac{c}{b}(a-2bc).$$ (this is a standard formular, for a parabola $ax^2+bx+c$ with $a>$ the minimum is attained at $-b/2a$).

Because of $z=x^2$ we can't allow negative values of $z$ so in the case $a < 2bc$ the minimum is at $z=0$, hence $x=0$. Otherwise we have $$x= \pm \sqrt{z_0}= \pm \sqrt{\frac{c}{b}(a-2bc)}.$$ If we plug $z_0$ into the parabola we get $$4bc(a-bc)$$ as the global minimum of our function in case $a \ge 2bc$.

In this last step I miscalculated last night. But now it fits.