Consider $J\colon X\rightarrow (-\infty,\infty]$ given by
$$J(x)=\dfrac{1}{2}a(x,x)+j(x)-(f,x)_{X},$$
where $a$ is bilinear form and $j\colon X\rightarrow (-\infty, \infty]$ is convex, proper and l.s.c. ($X$ - Hilbert space). I have the following inequality
$$J(x)\geq\dfrac{1}{2}m\|x\|^{2}-\|\alpha\|\|x\|-\|f\|\|x\|+\beta,\qquad (1)$$ for some $\alpha \in X$ and $\beta \in \mathbb{R}$. Now, I take a infimal sequance $(u_{n})$ i.e., $J(u_{n})\rightarrow l=\inf_{x\in X}(J(x))$. The question is: why is $(u_{n})$ bounded?
I think the only way to show that is by contradiction. It is easy to observe that $l<\infty$ because $J$ is proper. Assume that $(u_{n})$ is not bounded. We can choose a subsequance $(u_{n_{k}})$ such that $\|u_{n_{k}}\|\rightarrow \infty$. By the latter inequality ($m>0$) we obtain $J(u_{n_{k}})\rightarrow \infty$. Does it imply that $l=\infty$ (which leads to contradiction)?
$(u_{n_{k}})$ is only a subsequance of infimal sequance. Is $(u_{n_{k}})$ a infimal sequance either?
Recall Cauchy's inequality, which says that $$ |ab| \le \frac{a^2}{2} + \frac{b^2}{2} $$ for all $a,b \in \mathbb{R}$. For any $\epsilon >0$ we can make the following useful variant: $$ |ab | = \left\vert \sqrt{2\epsilon} a \frac{b}{\sqrt{2\epsilon}} \right\vert \le \frac{2\epsilon a^2}{2} + \frac{b^2}{2\cdot 2\epsilon} = \epsilon a^2 + \frac{b^2}{4 \epsilon}. $$
We use this as follows with $\epsilon = m/8$: $$ \Vert \alpha \Vert \Vert x \Vert \le \frac{m}{8} \Vert x\Vert^2 + \frac{2}{m}\Vert \alpha \Vert^2\qquad (2) $$ and $$ \Vert f \Vert \Vert x \Vert \le \frac{m}{8} \Vert x\Vert^2 + \frac{2}{m}\Vert f \Vert^2.\qquad (3) $$ Plugging this into your bound for $J$ reveals that $$ J(x) \ge \frac{m}{2} \Vert x\Vert^2 - \frac{m}{8} \Vert x\Vert^2 - \frac{2}{m} \Vert \alpha \Vert^2 -\frac{m}{8} \Vert x\Vert^2 - \frac{2}{m}\Vert f \Vert^2 - |\beta| \\ \ge \frac{m}{4} \Vert x \Vert^2 - \left(\frac{2}{m} \Vert \alpha \Vert^2 + \frac{2}{m}\Vert f \Vert^2 + |\beta| \right) =: \frac{m}{4} \Vert x \Vert^2 - \gamma, $$ where now $0\le \gamma \in \mathbb{R}$ is just some constant.
This bound now implies that if $\{x_n\}_n \subset X$ is any unbounded sequence, then $\{J(x_n)\}_n$ is unbounded as well. Indeed, $$ \frac{4}{m} (J(x_n) + \gamma) \ge \Vert x_n\Vert \Rightarrow \limsup_{n} \frac{4}{m}(J(x_n) +\gamma) =\infty \Rightarrow \limsup_n J(x_n) = \infty. $$
Now, for your minimizing sequence $\{u_n\}_n$, we know that $J(u_n) \to \inf_X J < \infty$. Thus the above shows that $\{u_n\}_n$ must be bounded, since otherwise $\{J(u_n)\}_n$ is unbounded, a contradiction.