Exercise: Prove the observability inequality for 1d wave equation

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I'm trying to prove the "observability inequality" for the 1d wave equation initial boundary value problem. To be more specific:

Let the initial boundary value problem

  • $\\u_{tt}-u_{xx}=0\;\;\forall x\in (0,1)\;\;and\;\forall t\in (0,T)\\u(x,0)=g(x)\;\;\forall x\in [0,1]\;\\u_t(x,0)=h(x)\;\;\forall x\in [0,1]\;\\u(0,t)=u(1,t)=0\;\;\forall t\in [0,T]\\$

Prove that $\frac {1}{2} \int_{0}^T \vert {u_x}^2(1,t)\vert \;dt \;\ge (T-2)E(0)\;\;for\;T \gt 2\;\;$ where $E(t)=\frac {1}{2} \int_{0}^1 {u_x}^2(x,t)\;+\;{u_t}^2(x,t)\;\;dx\;\;$ is the energy of the system.

The professor also gave us this hint: Use the multiplier method

So, I multilplied both sides of the initial equation by $\;xu_x\;$ and then I integrated the new equation in $\;[0,1]x[0,T]\;$. After calculations I concluded to:

$\frac {1}{2} \int_{0}^T {u_x}^2(1,t)\;dt\;=-\int_{0}^T \int_{0}^1 {u_x}^2\;dx\;dt\;\;+\int_{0}^1 xu_x(x,T)u_t(x,T)-xu_x(x,0)u_t(x,0) dx\;\;+\;\;TE(0)$

At this point I've been stuck for hours, I don't know how to proceed in order to prove the inequality. I assume that I should find a bound for the integral but I can't come up with any idea.

I would appreciate any hint or help. Thanks in advance!!