This inequality cannot be true. Let us rewrite it in the more common form $$P(R_n\ge x)\le e^{-x^2/2} \tag{1} $$ for $x\ge0$, where $R_n:=S_n/B_n$, $S_n:=\sum_1^n c_iB_i$, $B_n:=\sqrt{\sum_1^n c_i^2}$. Let $n=2$, $c_1=1$, and $c_2:=aI\{B_1=-1\}$, where $I\{\cdot\}$ denotes the indicator function, $a>0$ is large enough so that $\frac{-1+a}{\sqrt{1+a^2}}>x$, $x$ is less than $1$ but close enough to $1$ so that $e^{-x^2/2}<0.7$ (note that $e^{-1/2}=0.606\ldots<0.7$). Then $$P(R_2\ge x)=P(R_2\ge x,B_1=1)+P(R_2\ge x,B_1=-1)$$ $$=P(R_1\ge x,B_1=1)+P\Big(\frac{-1+aB_2}{\sqrt{1+a^2}}\ge x,B_1=-1\Big) $$ $$=P(B_1\ge x,B_1=1)+P(B_2=1,B_1=-1)$$ $$=P(B_1=1)+P(B_2=1,B_1=-1)$$ $$=\tfrac12+\tfrac14=0.75>0.7>e^{-x^2/2}, $$ so that $(1)$ fails to hold. Letting $c_3=\dots=c_n=0$, one disproves the inequality in question for any natural $n\ge2$. (What is sometimes referred to as the Azuma (or Hoeffding--Azuma) inequality is due entirely to **[[Hoeffding 1963][1]]**; see the last paragraph of Section 2 there.) [1]: http://www.tandfonline.com/doi/abs/10.1080/01621459.1963.10500830