For \(0<c<b<a\), let \((a+b-2 c) x^2+(b+c-2 a) x+(c+a-2 b)=0\) and \(\alpha \neq 1\) be one of its root. Then, …

Given

\((a+b-2c)x^2+(b+c-2a)x+(c+a-2b)=0\)

Observe that

\((a+b-2c)+(b+c-2a)+(c+a-2b)=0\)

Hence \(x=1\) is always a root.

Since \(\alpha\ne1\) is the other root,

\(\alpha=\dfrac{c+a-2b}{a+b-2c}\).

Let

\(p=a-b>0,\quad q=b-c>0\).

Then

\(a=b+p,\quad c=b-q\).

Substituting,

\(\alpha=\dfrac{(b-q)+(b+p)-2b}{(b+p)+b-2(b-q)}
=\dfrac{p-q}{p+2q}\).

Since \(p,q>0\),

\(-1<\dfrac{p-q}{p+2q}<1\).

Thus \(\alpha\in(-1,1)\).

Now let \(b\) be the geometric mean of \(a\) and \(c\):

\(b^2=ac=(b+p)(b-q)\).

This gives

\(b(p-q)=pq\).

Since \(b>0\),

\(p-q=\dfrac{pq}{b}>0\).

Hence

\(\alpha=\dfrac{p-q}{p+2q}>0\).

Therefore, if \(b\) is the geometric mean of \(a\) and \(c\), then necessarily

\(\alpha\in(0,1)\).

So:

(I) If \(\alpha\in(-1,0)\), then \(b\) cannot be the geometric mean of \(a\) and \(c\). True.

(II) If \(\alpha\in(0,1)\), then \(b\) may be the geometric mean of \(a\) and \(c\). True.

For example, take \(a=4,\ b=2,\ c=1\) \((b=\sqrt{ac})\).

Then

\(\alpha=\dfrac{4+1-4}{4+2-2}
=\dfrac14\in(0,1)\).

Hence both statements are true