If $x^2-bx+c=0$ has real roots, then prove that both are greater than $1$ when $c+1>b>2$. 
If    $ x^2-bx+c=0$ has real roots, prove that both roots are greater than $1$, when $c+1>b>2$.

Working
I tried to prove the given inequality by taking roots greater than $1$.
Let $\alpha$, $\beta$ be the roots of the quadratic equation. So $$\alpha+\beta =b$$
$$\alpha\cdot\beta =c$$
Since $\alpha>1$ and $\beta>1$, it can be deduced that,
$$\alpha+\beta >2\implies b>2$$ $$\alpha\cdot\beta >1\implies c>1\implies c+1>2$$
to combine these two inequalities I need another link between $c$ and $b$. How to proceed? Thanks.
 A: Given $x^2-bx+c=0$ we need


*

*$\Delta =b^2-4c\ge0$


and


*

*$x_1=\frac{b-\sqrt{b^2-4c}}{2}>1 \iff b-2> \sqrt{b^2-4c} \stackrel{b>2}\iff b^2-4b+4\ge b^2-4c \iff 4c+4\ge4b \iff c+1\ge b$

A: If $x^2-bx+c$ has both roots larger than $1$, replacing $x$ by $1$ should give the same sign as very large negative $x$, i.e. positive. Therefore we need $1-b+c>0$, which can be rearranged into the second inequality.
On the other hand, if $1-b+c>0$, and the roots are real, then both roots are either smaller than $1$ or larger than $1$. If both roots are smaller than $1$, obviously their sum is smaller than $2$, i.e. $b<2$, whereas if they are both larger than $1$, their sum is larger than $2$.
A: The roots of
$x^2-bx+c=0$
are
$x
=\dfrac{b\pm\sqrt{b^2-4c}}{2}
$.
If
$c+1 > b > 2$,
the smallest root is
$\dfrac{b-\sqrt{b^2-4c}}{2}
$
and
$b^2 > 4c$
so
we want
$b-\sqrt{b^2-4c}
\gt 2$
or
$(b-2)^2
\gt b^2-4c
$
or
$b^2-4b+4
\gt b^2-4c
$
or
$c+1 > b$
which we are given.
A: Needless to calculate the roots.
Set $\;p(x)=x^2-bx+c$.  If  $p(x)$ has real roots and $c+1>b>2$, $\; p(1)=1-b+c>0$, hence $1$ does not separate the roots, i.e. both are smaller or greater than $1$.
Now they're smaller or greater than $1$ exactly when their arithmetic mean  is, and this arithmetic mean is $\frac b2>1$ by hypothesis. The conclusion follows.
A: $$\begin{array}{rccccc}
& c + 1 & > & b & > & 2 \\
\implies & \alpha\beta+1 & > & \alpha+\beta & > & 2 \\
\implies & \alpha\beta-\alpha-\beta+1 & > & 0 & > & 2-\alpha-\beta \\
\implies & (\alpha-1)(\beta-1) & \underbrace{>}_{(\star)} & 0 & \underbrace{>}_{(\star\star)} & -\left(\;(\alpha-1)+(\beta-1)\;\right) \\
\end{array}$$
Now, $(\star)$ implies that $\alpha-1$ and $\beta-1$ have the same sign, while $(\star\star)$ implies that that sign must be positive. Thus, $\alpha> 1$ and $\beta > 1$. $\square$
