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Let : $$\alpha_{\text{min}} = \min \lbrace \alpha_{i}, \; 1 \leq i \leq n \rbrace.$$ Then, for all $x > 0$ and for all $i \in \lbrace 1,\ldots,n \rbrace$, $e^{-\alpha_{i}x} \leq e^{-\alpha_{\text{min}}x}$. Therefore : $$\forall x > 0, \; f(x) = \Big( \sum_{j=1}^{N} e^{-\alpha_{j}x} \Big)^{1/2} \leq \sqrt{N}e^{-\alpha_{\text{min}}x/2}.$$

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Let $\chi:\Bbb{F}_q\to\{\pm1\}$ be the canonical additive character $$\chi(x)=(-1)^{tr(x)}.$$ Fix an element $\ell\in\Bbb{F}_q^*$. Consider the function $f(x)=(1-\chi(\ell x))(1-\chi(\ell x^{-1}))$. It follows that for $a\in\Bbb{F}_q^*$ we have $f(a)=4$, if $tr(\ell a)=1=tr(\ell a^{-1})$ and $f(a)=0$ otherwise. Let $N(\ell)$ be the number of elements $a$ ...

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Yes you can. Rewrite $x$ as $\dfrac1{1+e^{z_2-z_1}}=\dfrac1{1+w}$. Then similarly $$y=\dfrac1{1+\sqrt[T]w}=\frac1{1+\sqrt[T]{\frac1x-1}}.$$

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