# Why is the gaussian part of this Bernstein-type inequality not trivial?

The following Bernstein-type inequality can be found in Introduction to the non-asymptotic analysis of random matrices.

Theorem Let $$X_1,\ldots X_n$$ be mean-zero sub-exponential random variables with $$\Vert X_i \Vert \leq K$$, then for any $$a \in \mathbb{R}^n$$ $$\mathbb{P}(\sum_{i=1}^n a_i X_i > t) \leq \exp(-c\min(\frac{t^2}{K^2\Vert a\Vert_2^2 }, \frac{t}{K\Vert a \Vert_\infty}))$$

Proof By appropriate rescaling of $$X_i$$ and $$t$$ we may assume that $$K = 1$$. Now the markov inequality yields that for any $$\lambda > 0$$ \begin{align*}\mathbb{P}(\sum_{i=1}^n a_i X_i > t) &= \mathbb{P}(\exp(\lambda \sum_{i=1}^n a_i X_i) > \exp(\lambda t))\\ &\leq \exp(-\lambda t)\mathbb{E}(\exp(\lambda\sum_{i=1}^n a_i X_i))\\ &= \exp(-\lambda t) \prod_{i=1}^n\mathbb{E}\exp(\lambda a_i X_i)\\ &\leq \exp(-\lambda t) \prod_{i=1}^n \exp(C\lambda^2 a_i^2) \qquad \forall \lambda < \frac{C}{\Vert a\Vert_\infty } \end{align*} where the last inequality uses an equivalent definition of subexponential random variables. Finally, minimalisation of the second order polynomial $$-\lambda t + C \lambda^2 \Vert a \Vert_2^2$$ for $$\lambda <\frac{C}{\Vert a \Vert_\infty}$$ yields the result. $$\square$$

My confusion is due to the minimalisation of this second order polynomial. If I understand rightly the optimisation uses $$\lambda = \frac{t}{2C\Vert a \Vert_2^2}$$ if this is possible but the boundary value otherwise which is why we have to put a minimum.

Why can't we just use the boundary value instead of adding a minimum? It appears to me that this would yield a stronger result.