Cleverly finding the minimum of function

Question

I want to cleverly find minimum of function $$f(x,y)=a(y-x)^2+bx^2$$ when $$x^2+y^2=1.$$ By cleverly I mean by using some smart inequalities which give lower estimate for $f,$ and which become equalities exactly when the minimum occurs. Thus, by cleverly, I do not mean Lagrange's multipliers method/finding zero of the first derivative. By doing the former, I have obtained $$\min{f}=a\bigg[1-{1\over 2}\bigg[\sqrt{4+{b^2 \over a^2}}-{b \over a} \bigg]\bigg],$$ which is attained for $$x^2={{4+{b^2 \over a^2}-{b \over a}\sqrt{4+{b^2 \over a^2}}}\over{2(4+{b^2 \over a^2}})}.$$

Answer 1

Let $x=\cos{t},y=\sin{t}$, where $t\in [0.2\pi)$, then we have

$\begin{array}\\ f&=a-2axy+bx^2\\ &=a-2a\sin{t}\cos{t}+b\cos^2{t}\\ &=-a\sin{2t}+\displaystyle\frac{b}{2}\cos{2t}+a+\frac{b}{2}\\ &=\displaystyle\sqrt{\frac{b^2}{4}+a^2}\sin(2t+\phi)+a+\frac{b}{2}\\ &\geq-\displaystyle\sqrt{\frac{b^2}{4}+a^2}+a+\frac{b}{2}\\ \end{array} $

where $\displaystyle\phi=\arctan(\frac{b}{-2a})\in[0,2\pi)$.

Answer 2

We consider only the case $a,b>0$, of course. Then after rescaling, we can and do assume $a,b$ are of the shape: $$ \begin{aligned} b &= \sin c\ ,\\ a &= \frac 12\cos c\ . \end{aligned} $$ In the sequel, $c$ is thus a constant in the interval between $0$ and $\pi/2$. The variables $x,y$, constrained to $x^2+y^2=1$ are then parametrized by an unconstrained $t\in[0,2\pi)$ with $$ \begin{aligned} x&=\cos t\ \\ y&=\sin t\ . \end{aligned} $$ So we need to minimize: $$ \begin{aligned} f(t) &=\cos c\frac 12(\sin t -\cos t)^2 +\sin c\cos^2 t \\ &=\cos c\left(\cos t\cos\frac \pi4 -\sin t\sin\frac \pi4 \right)^2 +\sin c\cos^2 t \\ &=\cos c\cos\left(t+\frac \pi4 \right)^2 +\sin c\cos^2 t \\ &=\cos c\frac 12\left(\cos2\left(t+\frac \pi4 \right)+1\right) +\sin c\frac 12(\cos2 t+1) \\ &=\frac 12\cos c\cos\left(2t+\frac \pi2 \right) +\frac 12\sin c\cos2 t +\text{Constant} \\ &= -\frac 12\cos c\sin2t +\frac 12\sin c\cos2 t +\text{Constant} \\ &= \frac 12\sin (c-2 t) +\text{Constant}\ . \end{aligned} $$ Now we have to minimize the sine part.

Answer 3

$\begin{array}\\ f(x,y) &=a(y-x)^2+bx^2\\ &=ay^2-2axy+ax^2+bx^2\\ &=a(1-x^2)-2axy+ax^2+bx^2\\ &=a-2axy+bx^2\\ &=a-2ax\sqrt{1-x^2}+bx^2\\ f_x(x, y) &=-2a(\sqrt{1-x^2}-\dfrac{x^2}{\sqrt{1-x^2}})+2bx\\ &=-2a(\dfrac{1-x^2-x^2}{\sqrt{1-x^2}})+2bx\\ &=-2a(\dfrac{1-2x^2}{\sqrt{1-x^2}})+2bx\\ &=0 \end{array} $

when (since I don't see any clever inequalities I'll use brute force)

$a(\dfrac{1-2x^2}{\sqrt{1-x^2}})=bx $ or $a(1-2x^2) =bx\sqrt{1-x^2} $ or $a^2(1-4x^2+4x^4) =b^2x^2(1-x^2) =b^2x^2-b^2x^4 $ or $x^4(4a^2+b^2)-x^2(4a^2+b^2)+a^2 =0$.

Therefore

$\begin{array}\\ x^2 &=\dfrac{4a^2+b^2\pm\sqrt{(4a^2+b^2)^2-4a^2(4a^2+b^2)}}{2(4a^2+b^2)}\\ &=\dfrac{4a^2+b^2\pm\sqrt{16a^4+8a^2b^2+b^4-16a^4-4a^2b^2}}{2(4a^2+b^2)}\\ &=\dfrac{4a^2+b^2\pm\sqrt{4a^2b^2+b^4}}{2(4a^2+b^2)}\\ &=\dfrac{4a^2+b^2\pm b\sqrt{4a^2+b^2}}{2(4a^2+b^2)}\\ &=\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}}\\ \text{so}\\ y^2 &=1-x^2\\ &=\dfrac12\mp\dfrac{b}{2\sqrt{4a^2+b^2}}\\ \text{and}\\ f(x, y) &=a-2a\sqrt{\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}}}\sqrt{\dfrac12\mp\dfrac{b}{2\sqrt{4a^2+b^2}}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a-2a\sqrt{\dfrac14-\dfrac{b^2}{4(4a^2+b^2)}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a-a\sqrt{1-\dfrac{b^2}{4a^2+b^2}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a-a\sqrt{\dfrac{4a^2+b^2-b^2}{4a^2+b^2}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a-a\sqrt{\dfrac{4a^2}{4a^2+b^2}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a-2a^2\sqrt{\dfrac{1}{4a^2+b^2}}+b(\dfrac12\pm\dfrac{b}{2\sqrt{4a^2+b^2}})\\ &=a+\dfrac{b}{2}-\dfrac{2a^2}{\sqrt{4a^2+b^2}}\pm(\dfrac{b^2}{2\sqrt{4a^2+b^2}})\\ &=a+\dfrac{b}{2}-\dfrac{4a^2\mp b^2}{2\sqrt{4a^2+b^2}}\\ \end{array} $

If we want the minimum value, take the lower sign, so the min is $a+\dfrac{b}{2}-\dfrac{4a^2+ b^2}{2\sqrt{4a^2+b^2}} =a+\dfrac{b}{2}-\frac12\sqrt{4a^2+b^2} =a+\dfrac{b}{2}-\sqrt{a^2+\dfrac{b^2}{4}} $.