<s>[INST] 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.
 [/INST] 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.
</s>