Cleverly finding the minimum of function 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 $$x^2={{4+{b^2 \over a^2}-{b \over a}\sqrt{4+{b^2 \over a^2}}}\over{2(4+{b^2 \over a^2}})}.$$

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