Three quadratic equations have positive roots The three quadratic equations $ax^2-2bx+c=0$, $bx^2-2cx+a=0$ and $cx^2-2ax+b=0$ has both roots positive. Then which of the following is/are true
A) $a^2=bc$
B) $b^2=ac$
C) $c^2=ab$
D) $a=b=c$
I assumed $a>0$ so parabola is open upwards and hence it should definitely cut positive $Y $axis as it has both roots positive and hence $c \gt 0$.
Since $c \gt 0$ the third parabola is open upwards and so $b \gt 0$ and finally with same reasoning $a \gt 0$. Now each Discriminant is Non negative so
$$b^2 \ge ac$$
$$c^2 \ge ab$$
$$a^2 \ge bc$$
adding all
$$a^2+b^2+c^2-ab-bc-ca \ge 0$$ $\implies$
$$(a-b)^2+(b-c)^2+(c-a)^2 \ge 0$$
if we take equality above i will get $a=b=c$ so fourth option is valid. But how to check whether other options a is/are true
 A: Since the equations have two roots, the leading coefficients must be non-$0\,$, so $abc \ne 0$. Assume WLOG that $a \gt 0\,$ (otherwise use the same argument for $a,b,c \mapsto -a,-b,-c\,$). Since both roots of each equation are positive, it follows by Vieta's relations that $b \gt 0 , c \gt 0\,$ as well, so in the end all three of $a,b,c$ are strictly positive numbers. Now, picking up from where the OP left off: 

$$b^2 \ge ac$$
  $$c^2 \ge ab$$
  $$a^2 \ge bc$$

Those are all positive numbers, then multiplying together gives $\,a^2 b^2 c^2 \ge a^2 b^2 c^2\,$. But the latter is in fact an equality, which means that each of the previous three inequalities must be an equality as well, therefore $\,a^2=bc, b^2=ca, c^2=ab\,$. Furthermore, $a^2=bc=\sqrt{ac} \cdot c \implies a=c\,$, so in the end $a=b=c$.
A: I would like to warn you that the condition you've got, i.e.
$$
(a-b)^2+(b-c)^2+(c-a)^2 \ge 0
$$
is a necessary condition and not a sufficient one. It means that the testing of D against it just ensures that $a,b,c$ may satisfy D, but does not ensure that they do. Moreover, it is pretty useless condition as a sum of squares is always non-negative for real numbers, so it has, in fact, no information about $a,b,c$ at all.
We can start by saying that for the three equations to have two root each we need to have $a,b,c$ nonzero. Then we continue by writing down all three solutions
\begin{align}
&\frac{b}{a}\pm\frac{\sqrt{b^2-ac}}{a}\ge 0,\\
&\frac{c}{b}\pm\frac{\sqrt{c^2-ab}}{b}\ge 0,\\
&\frac{a}{c}\pm\frac{\sqrt{a^2-bc}}{c}\ge 0.
\end{align}
Clearly $\frac{b}{a}>0$, $\frac{c}{b}>0$ and $\frac{a}{c}>0$, that is $a,b,c$ have the same sign. Since the roots does not change after multiplying the equations by an arbitrary number, we can assume WLOG that  $a,b,c>0$. Hence, we know that
\begin{align}
b^2-ac\ge 0\quad\Leftrightarrow\quad b^3\ge abc,\\
c^2-ab\ge 0\quad\Leftrightarrow\quad c^3\ge abc,\\
a^2-bc\ge 0\quad\Leftrightarrow\quad a^3\ge abc.\\
\end{align}
Adding together and dividing by $3$ we get
$$
\frac{a^3+b^3+c^3}{3}\ge abc=\sqrt[3]{a^3b^3c^3}.
$$
The AM-GM inequality gives then that it is, in fact, equality, therefore $a^3=b^3=c^3$, which implies $a=b=c$, so D is true.
It makes A,B,C to be trivially true as well.
