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.

Given $x^2-bx+c=0$ we need


*

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


and


*

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

