<s>[INST] 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
$$x^2={{4+{b^2 \over a^2}-{b \over a}\sqrt{4+{b^2 \over a^2}}}\over{2(4+{b^2 \over a^2}})}.$$
 [/INST] 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.
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