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Please, can somebody help me with this problem? I tried to use the Mean Value Theorem, but couldn't solve it.

Let $g: [a,b]\rightarrow\mathbb{R}$ a differentiable function on $[a,b]$. If $g^{\prime}(\theta)\neq 0$, then given $\alpha\in(0,1)$, there's $\delta > 0$ such that $$|g(x) - g(y)|^2\ \geq\ 2(1-\alpha)^2|g^{\prime}(\theta)|^2\left|{\atop}|x-\theta| - |y-\theta|\right|^2,$$ for all $x,y\in(\theta-\delta,\theta+\delta)$.

Thanks in advance
FASCH
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    Is your $\alpha$ fixed beforehand? If you take $g(s)=s$, $\theta =0$, $x>y>0$, then after simplifications you arrive to $1\ge 2(1-\alpha)^2$, which is not always true. – TZakrevskiy Jul 20 '13 at 20:30

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