Upper bound on a set of functions

Question

Consider a finite set of functions $\{f_1(x),f_2(x),...f_n(x)\}$ where $f_i: \mathbb{R}_+ \to \mathbb{R}_+, \forall i$. Suppose there exist $a_i\ge0$ and $b_i > 0$ such that \begin{align} f_i(x) \le e^{a_i - b_i x}, \qquad \forall i, \forall x\ge0. \end{align} Let us define the following three scalars: \begin{align} a:=\max_i\{a_i\} \qquad T > \max_i\{ \frac{a_i}{b_i} \} \qquad b:=\min_i\{ b_i - \frac{a_i}{T} \}. \end{align} Prove that \begin{align} f_i(x) \le e^{ \frac{a}{2} - bx } , \qquad \forall i, \forall x \ge0. \end{align}

Answer 1

This is false. Let $i = 1$, $a_1 = b_1 = 1$, and let: $$ f_1(x) = e^{a_1 - b_1x} = e^{1-x} $$ Then: $$ a = a_1 \\ T = \frac{a_1}{b_1} \\ b = b_1 - \frac{a_1}{\frac{a_1}{b_1}} = 0 $$ So the inequality asserts that: $$ f_1(x) = e^{1-x} \leq e^{\frac{1}{2}} \; \forall i, \, \forall x \geq 0 $$ However, this is not true, as $f_1(0) = e > e^{\frac{1}{2}}$.