<s>[INST] Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdot \:x_2}{2}\right)^2$
$=\:x_1\cdot \:x_2\le \:\frac{\left(x_1\cdot \:\:x_2\right)^2}{4}\:$
$=\:\:x_1\cdot \:\:x_2\le \:\:\frac{{x_1}^2+2x_1x_2+{x_2}^2}{4}$
$=4\left(x_1\cdot x_2\right)\:\le {x_1}^2+2x_1x_2+{x_2}^2$
$=4\left(x_1\cdot \:x_2\right)\:-2x_1x_2\le \:{x_1}^2+{x_2}^2$
Substituting in the values for $x_1$ and $x_2$ we get:
$4\left(\frac{a+\sqrt{a^2-4a+4}}{2}\cdot \:\frac{a-\sqrt{a^2-4a+4}}{2}\right)\:-2\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)\le \:\left(\frac{a+\sqrt{a^2-4a+4}}{2}\right)^2+\left(\frac{a-\sqrt{a^2-4a+4}}{2}\right)^2$
$= 4\left(a-1\right)\:-\left(2a-2\right)\le \:a^2-2a+2$
$ = 0\le a^2-4a+4$
It seems as if I walked in circles through this process, can anyone help? Thanks in advance!
 [/INST] you have that:
$$(x-x_1)(x-x_2)=0$$
since these are the roots, so:
$$x^2-(x_1+x_2)x+x_1x_2=0$$
so:
$$x_1+x_2=a\qquad x_1x_2=a-1$$
now we can say:
$$(x_1+x_2)^2=x_1^2+2x_1x_2+x_2^2=(x_1^2+x_2^2)+2(a-1)=a^2$$
which finally gives us:
$$y=x_1^2+x_2^2=a^2-2a+2$$
now to find the minimum differentiate:
$$\frac{dy}{da}=2a-2$$
so for $y'=0$ we get $a=1$ giving $\min(x_1^2+x_2^2)=1$

In case you are not well versed on the use of derivatives, here is an alternative:
$$\begin{align}y=&a^2-2a+2\\
=&(a-1)^2+1\end{align}$$
from this we can see that the minimum of the function occurs where $(a-1)=0\Rightarrow a=1$

Also, here is a graph to visualise what is going on
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