The ineq can be reframed as $c_1,c_2,\cdots,c_{n-1}\le 0,c_1+c_2+\cdots c_n=0$ Then $$\sum_{i=1}^n e^{c_i}-e^{-c_i}\ge 0$$. We see that $f(x)=e^x-e^{-x}$ is concave in $(-\infty,0])$. Now note that $$(0,c_1,\cdots c_{n-1})\prec (0,0,\cdots,c_1+c_2+\cdots c_{n-1})$$. Hence by karamata, $\sum_{i=1}^{n-1}f(c_i)\ge f(c_1+c_2+\cdots c_{n-1})=f(-c_n)$. Hence ineq to be proved is $f(-c_n)+f(c_n)\ge 0$ which is infact an equality.