We have $10$ numbers in the interval $(0,1)$, not necessarily distinct. At any moment, we can choose two of them, $a$ and $b$. If the quadratic $x^2-ax+b$ has two (possibly identical) real roots, we can replace $a$ and $b$ by the two roots. Can this process go on forever?
One interesting case is when $a=b$. The polynomial is $x^2-ax+a$, which has roots $\dfrac{a\pm\sqrt{a^2-4a}}{2}$, but when $a\in(0,1)$, this is not a real number.
-First note that in order for this to have real roots we need to have $a^2-4b >0$, which for $a,b \in (0,1)$ implies $a > b$. In fact it implies $a >4b$, but I don't think I need that.
-Next note that if a particular choice of $a,b$ has real roots then they are strictly between $b$ and $a$.
-This means that at any given step the minimum value of our $10$ numbers can only increase or stay the same. Let's call the initial smallest value $b_0 >0$.
-Now consider the sum of our $10$ numbers. If we replace two numbers $a,b$ in by this process, this sum gets decreased by $b$. In particular it gets decreased by at least $b_0$, but it also needs to always be positive. Therefore we see that this process cannot go on forever.