How prove that there exists $m$ such that $|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$ Let the function $$f(x)=x^2+bx+c,\qquad b^2-4c>0$$ Assume that $x_{1},x_{2}$ are the roots of $f(x)$ and $|x_{1}-x_{2}|\ge 1$.

Show that: There exists an integer $m$, such that
  $$|f(m)|\le \dfrac{\sqrt{b^2-4c}}{2}$$

My idea: Since $$x_{1}+x_{2}=-b,x_{1}x_{2}=c$$
and $$f(x)=(x-x_{1})(x-x_{2})$$
so
$$|x_{1}-x_{2}|=\sqrt{(x_{1}+x_{2})^2-4x_{1}x_{2}}=\sqrt{b^2-4c}$$
so
$$|f(m)|=|(m-x_{1})(m-x_{2})|\le \dfrac{(x_{1}-x_{2})^2}{4}$$
Then I can't.Thank you very much.
 A: Without loss of generality we may suppose that $x_1<x_2$. Let us consider two cases:


*

*Case 1: $\Delta~\buildrel{\rm def}\over{=}~b^2-4c\leq 4$. Since $|x_1-x_2| \geq 1$, there is an integer $m$ that belongs to the interval $[x_1,x_2]$. Now
$$0\geq f(m)\geq f\left(-\frac{b}{2}\right)=-\frac{\Delta}{4}\geq-\frac{\sqrt{\Delta}}{2}$$
and the desired inequality is satisfied in this case.

*Case 2: $\Delta>4$. In this the function $t\mapsto f(t)$ is strictly increasing on $[-b/2,+\infty)$. The equation $f(x)=-\frac{\sqrt{\Delta}}{2}$ has a unique solution $y\in [-b/2,+\infty)$ given by
$$y=\frac{-b+\sqrt{\Delta-2\sqrt{\Delta}}}{2}$$
and similarly The equation $f(x)=\frac{\sqrt{\Delta}}{2}$ has a unique solution $z\in [-b/2,+\infty)$ given by
$$z=\frac{-b+\sqrt{\Delta+2\sqrt{\Delta}}}{2}$$
Now
$$\eqalign{z-y&=\frac{\sqrt{\Delta+2\sqrt{\Delta}}-\sqrt{\Delta-2\sqrt{\Delta}}}{2}\cr
&=\frac{2}{\sqrt{1+\frac{2}{\sqrt{\Delta}}}+\sqrt{1-\frac{2}{\sqrt{\Delta}}}}>1
}$$
here we used the easy inequality : $\dfrac{2}{\sqrt{1+u}+\sqrt{1-u}}>1$. Now, since $z-y>1$ there is an integer $m$ that belongs to $[y,z]$ but then
$$-\frac{\sqrt{\Delta}}{2}=f(y)\leq f(m)\leq f(z)=\frac{\sqrt{\Delta}}{2}$$
and the desired conclusion follows.$\qquad\square$

A: We consider the following equivalent problem.
For $ d \geq 1 $, define $ g(x) = x^2 + dx$. We want to show that there exists real numbers $ j \leq 0 \leq k$ with $k-j \geq 1$ such that for all $ j \leq x \leq k$, we have
$$ - \frac{d}{2} \leq g(x) \leq \frac{d}{2} . $$
Proof of Equivalent: To see why this is equivalent, is because we can go from $g(x)$ to $f(x)$ by a horizontal translation, and the translated range of $[j,k]$ guarantees us that we have an integer value.
Proof of new problem: With this simplified version, we can easily solve the inequality.
For $ 1 \leq d \leq 2$, the solution set is
$$ \frac{1}{2} ( -d - \sqrt{d(d+2)}) \leq x \leq \frac{1}{2}( \sqrt{d(d+2)}-d).$$
Verify that $k-j = \sqrt{d(d+2)} \geq \sqrt{3} \geq 1 $.
For $ 2 < d$, the solution set (which contains 0) is
$$\frac{1}{2} ( \sqrt{ d(d-2)} -d) \leq x \leq \frac{1}{2} ( \sqrt{ d(d+2) } -d ) .$$
Verify that $ k-j = \frac{1}{2} ( \sqrt{ d(d+2) } - \sqrt{d(d-2)}) = \frac{2d}{ \sqrt{d^2+2d} + \sqrt{d^2+2d}} > \frac{2d}{2d} = \geq 1$.
