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!
 A: 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
A: 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 $a = 1$.
A: 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.
A: $x_1^2+x_2^2=(x_1+x_2)^2-2(x_1x_2)=a^2-2(a-1)$
The last equility is Vieta or if you wish $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+(x_1x_2).$
So the min is when $a=1$ (because of usual calculus).
