Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$ Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$

We can use the Vasc's EV Method.
See here: https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
Indeed, let $a+b+c+d+e=constant.$
Thus,  by corollary 1.9, case 1(b) ($p=0$,$q=-1$)
the expression $\sum\limits_{cyc}\frac{1}{a}$ gets a minimal value for equality case of four variables.
Id est, it remains to prove our inequality for $b=c=d=e$ and $a=\frac{1}{e^4}$, which gives
$$(e-1)^2(40e^8+80e^7+120e^6+160e^5-132e^4-89e^3-46e^2-3e+40)\geq0,$$
which is true because $$40e^8+80e^7+120e^6+160e^5-132e^4-89e^3-46e^2-3e+40=$$
$$=40(e^4+e^3+e^2-3e+1)^2+e(320e^4-12e^3+71e^2-486e+237)\geq$$
$$\geq e((e^2+40.5)(3e-2)^2+311e^4-297.5e^2+75)>0$$ and the last inequality is true because $$297.5^2-4\cdot311\cdot75=297.5^2-311\cdot300<0.$$
