<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] It looks like we are asked to notice that $ \ x^2 - ax + (a-1) \ = \ (x - 1)·( \ x - [a - 1] \ ) \ = \ 0 \ \ , $ so that one solution to the quadratic equation is always $ \ x_1 \ = \ 1 \ \ , $ with the other being $ \ x_2 \ = \ a - 1 \ \ . $  The function in question is then $ \ x_1^2 + x_2^2 \ = \ 1 + (a - 1)^2 \ \ , $ which will attain its minimum for $ \ a - 1 \ = \ 0 \ \Rightarrow \ a = 1 \ \ . $
As for sorting this out by the use of inequalities, that might work better with the RMS-GM inequality:
$$ \sqrt{\frac{x_1^2 \ + \ x_2^2}{2}} \ \ \ge \ \ \sqrt{x_1·x_2} \ \ \ge \ \ 0 \ \  \ \ \Rightarrow \ \ x_1^2 \ + \ x_2^2 \ \ \ge \ \ 2·x_1·x_2 \ \ = \ \  2·(a \ - 1)  \ \ \ge \ \ 0 \ \ , $$
where we have used the Viete relation for the product of the zeroes.
</s>