<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] Another way to go...  From $x^2 - a x + (a-1)$, we know that any root satisfies
$$  x^2 = ax - (a-1)  \text{,}  $$
so \begin{align*}
x_1^2 + x_2^2 &= (ax_1 - (a-1)) + (ax_2 - (a-1))  \\
    &= a (x_1 + x_2) - 2a + 2  \text{.}
\end{align*}
From \begin{align*}
x^2 -ax +(a-1) &= (x-x_1)(x-x_2)  \\
    &= x^2 -(x_1 + x_2)x + x_1x_2  \text{,}  \\
\end{align*}
we have $x_1 + x_2 = a$, so \begin{align*}
x_1^2 + x_2^2 &= a(a) - 2a + 2  \\
    &= a^2 - 2a + 1 - 1 + 2  \\
    &= (a-1)^2 + 1  \text{,}  
\end{align*}
which is a nonnegative term plus $1$, so is minimized when the nonnegative term is zero, that is, when $a = 1$.
</s>