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As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and  $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)> t)dt=\\ \int_1^\infty\mathbb{P}( Z> c\sqrt{\log t})dt\leqslant \int_1^\infty \min(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1, $$ as we need.

Explanations: we estimated the probability $\mathbb{P}( Z> c\sqrt{\log t})$ from above as $2\exp(-3\log t)$, this is equivalent to $$\mathbb{P}\left(Z>x\right)\leqslant 2\exp\left(-\frac{x^2}{2||a||^2_2}\right)$$ for $x=c\sqrt{\log t}$. Sometimes this is worse than the trivial upper estimate $\mathbb{P}( Z> c\sqrt{\log t})\leqslant 1$, that's why the minimum.

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and  $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)> t)dt=\\ \int_1^\infty\mathbb{P}( Z> c\sqrt{\log t})dt\leqslant \int_1^\infty \min(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1, $$ as we need.

Explanations: we estimated the probability $\mathbb{P}( Z> c\sqrt{\log t})$ from above as $2\exp(-3\log t)$, this is equivalent to $$\mathbb{P}\left(Z>x\right)\leqslant 2\exp\left(-\frac{x^2}{2||a||^2_2}\right)$$ for $x=c\sqrt{\log t}$. Sometimes this is worse than the trivial upper estimate $\mathbb{P}( Z> c\sqrt{\log t})\leqslant 1$, that's why the minimum.

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)\geqslant t)dt=\\ \int_1^\infty\mathbb{P}( Z\geqslant c\sqrt{\log t})dt\leqslant \int_1^\infty \max(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1, $$ as we need.

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)> t)dt=\\ \int_1^\infty\mathbb{P}( Z> c\sqrt{\log t})dt\leqslant \int_1^\infty \min(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1, $$ as we need.

Explanations: we estimated the probability $\mathbb{P}( Z> c\sqrt{\log t})$ from above as $2\exp(-3\log t)$, this is equivalent to $$\mathbb{P}\left(Z>x\right)\leqslant 2\exp\left(-\frac{x^2}{2||a||^2_2}\right)$$ for $x=c\sqrt{\log t}$. Sometimes this is worse than the trivial upper estimate $\mathbb{P}( Z> c\sqrt{\log t})\leqslant 1$, that's why the minimum.

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)\geqslant t)dt=\\ \int_1^\infty\mathbb{P}( Z\geqslant c\sqrt{\log t})dt\leqslant \int_0^\infty \max(1,2\exp(-3\log t))dt<1, $$ as we need.

As I understand these notations, $\Psi_2(x)=\exp(x^2)-1$ and $\|Z\|_{\Psi}=\inf \{c>0 : \mathbb{E} (\Psi(|Z| / c)) \leqslant 1\}$. So the inequality $\|Z\|_{\Psi}\leqslant c$ means that $\mathbb{E} (\Psi(|Z| / c)) \leqslant 1$. Denote $c=\sqrt{6}\|a\|_2$, $Z=|\sum_i \epsilon_i a_i|$. We have $$ \mathbb{E} (\Psi_2(Z/c))=\mathbb{E} \exp(Z^2/c^2)-1=-1+\int_0^\infty\mathbb{P} (\exp(Z^2/c^2)\geqslant t)dt=\\ \int_1^\infty\mathbb{P}( Z\geqslant c\sqrt{\log t})dt\leqslant \int_1^\infty \max(1,2\exp(-3\log t))dt<2\int_1^\infty t^{-3}dt=1, $$ as we need.