Let $a,b,c \in \mathcal{L}_2^n(\mathbb{R}_+)$, and let $\|\cdot\|_2$ be the norm in $\mathcal{L}_2^n(\mathbb{R}_+)$.
Suppose that there exists $\varepsilon > 0$ (independent of $a,b$ or $c$) such that $$\|a\|_2^2 - \|b\|_2^2 \leq -\varepsilon\|c\|_2^2.$$ Then, can I say that there exists $\tilde{\varepsilon} > 0$ (again independent of $a,b$ or $c$) such that $$\|a\|_2 - \|b\|_2 \leq -\tilde{\varepsilon}\|c\|_2$$ and if so, how could I prove it?
I tried using concavity of the square root function, as well as the fact that $\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}$, both to no success. Any help or indication in the direction of a proof (or a proof that this is false, for that matter) would be greatly appreciated.
EDIT:
To clarify, by $\mathcal{L}_2^n(\mathbb{R}_+)$ I mean $$\mathcal{L}_2^n(\mathbb{R}_+) = \left\{f:\mathbb{R}_+\to\mathbb{R}^n \;\middle|\; \int_0^\infty\!f(t)^Tf(t)\,dt < \infty\right\}$$
Just to add some more information, we can say that $a = f_1(c)$ and $b = f_2(c)$, with $f_1,f_2$ bounded operators from $\mathcal{L}_2^n(\mathbb{R}_+)$ into itself. Then, there exist $0 < \gamma_1 < \infty$ and $0 < \gamma_2 < \infty$ such that $\|a\| \leq \gamma_1\|c\|$ and $\|b\| \leq \gamma_2\|c\|$. Using this, I get that $$\|a\|_2^2 - \|b\|_2^2 \leq -\varepsilon_2\left(\|a\|_2^2 + \|b\|_2^2\right)$$ where $\varepsilon_2 = \min\left\{\frac{\varepsilon}{2\gamma_1^2},\frac{\varepsilon}{2\gamma_2^2}\right\}$. Then, after some manipulation, we have that $$\|a\|_2 - \|b\|_2 \leq -\varepsilon_3\left(\|a\|_2 + \|b\|_2\right)$$ with $\varepsilon_3 = \min\left\{\sqrt{1+\varepsilon_3} - 1,1 - \sqrt{1-\varepsilon_3}\right\}$. But then I don't know how to go back to $\|c\|$ to conclude...
You might want to be more specific with dependencies. But, without restrictions, you have $$ \|a\|_2^2-\|b\|_2^2=(\|a\|_2-\|b\|_2)(\|a\|_2+\|b\|_2). $$ So $$ \|a\|_2-\|b\|_2\leq- \left(\frac{\varepsilon\|c\|_2}{\|a\|_2+\|b\|_2} \right)\,\|c\|_2 $$