Let $(v_n)_n$, $(t_n)_n$ be two sequences such that $(v_n)_n$ is bounded in $H^1_0(\Omega)$, with $\Omega$ open bounded domain in $\mathbb{R}^2$ and $(t_n)_n$ is a real bounded sequence. Let $$u_n:= v_n +\alpha t_n \quad\mbox{ for all } n\in\mathbb{N},$$ with $\alpha\in\mathbb{R}^+$. During the math class, the professor said that the above relation ensures that, given $\varepsilon>0$, a constant $c(\varepsilon)>0$ exists such that: $$(1)\qquad\qquad u_n^2 \ge (1-\varepsilon)t_n^2 \alpha^2-c(\varepsilon) v_n^2,$$ $$(2)\qquad\qquad \lim_n\|u_n\|^2 \le (1+\varepsilon)\lim_n t_n^2,$$ where $\|\cdot\|$ denotes the norm in $H^1_0(\Omega)$.
Could someone please help me to understand why these 2 relations actually hold true? Thank you in advance.
${\bf EDIT:}$ For the inequality $(2)$ some changes are needed in my above writing. To prove that $(2)$ holds, in math class we referred to $u_n:= v_n+t_n w_n$, with $\|w_n\|=1$ (I apologize with psl2Z whom was trying with the wrong one sequence). I'd proceed in this way $$\|u_n\|^2 =\int_{\Omega}|\nabla (v_n+t_n w_n)|^2 dx\le \|v_n\|^2 +t_n^2.$$ Actually, I don’t even understand here the role of $\varepsilon$.
For (1): for every $\varepsilon > 0$: $-2\alpha t_n v_n \leq \varepsilon (\alpha t_n)^2 + \frac{1}{\varepsilon}v_n^2$, therefore $u_n^2 = (\alpha t_n)^2 + 2\alpha t_n v_n + v_n^2 \geq (1-\varepsilon)\alpha^2 t_n^2 - (\frac{1}{\varepsilon}-1)v_n^2$. If $\epsilon < 1$ then $C(\varepsilon) = \frac{1}{\varepsilon}-1 > 0$, otherwise the inequality is clear.