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:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$

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This is only a partial solution but I think someone more familiar with such elementary inequalities than myself might be able to finish it. You can replace $a,b,c,d$, and $e$ with their reciprocals and the inequality in question becomes
$$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}$$
Since still $abcde = 1$, we can rewrite this as
$$a + b + c + d + e + {33 \over 2}{1 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})} \geq {83 \over 10}(abcde)^{1 \over 5}$$
Some algebra converts this into
$${a + b + c + d + e \over 5} – (abcde)^{1 \over 5} \geq {33 \over 50}(abcde)^{1 \over 5} – {33 \over 50}{5 \over ({1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e})}$$
In other words, $AM – GM \geq {33 \over 50}(GM – HM)$. This is needed only when $abcde = 1$, but by scaling this should then hold for all $a,b,c,d,$ and $e$. So you inequality experts out there… is this something that follows from well-known inequalities?

Here is a way using “smoothing”. We need to show given $abcde=1$,
$$\frac1a + \frac1b + \frac1c + \frac1d +\frac1e+ \frac{33}{2(a + b + c + d+e)} \ge \frac{83}{10}$$

Consider replacing any two variables, WLOG $a, b$, with $\sqrt{ab}, \sqrt{ab}$. Clearly the constraint is maintained, and the RHS is untouched.

However, for the LHS, note that by AM-GM

$$\frac1a +\frac1b \ge \frac2{\sqrt{ab}}; \quad a+b \le 2\sqrt{ab}$$
so the LHS decreases. By continuing the process of smoothing, every variable gets replaced by the geometric mean, and the LHS only decreases, so it suffices to prove the inequality for $a=b=c=d=e=\sqrt[5]{abcde}=1$, which gives trivially an equality.

Let $$f(a,b,c,d,e)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}-{\frac{{83}}{10}}+\lambda(abcde-1).$$
Since $\lim\limits_{a\rightarrow0^+}f=+\infty$ and $\mathbb R_{+}^5$ is a compact,

$f$ gets a minimal value inside $\mathbb R_{+}^5$ and in the minimum point $(a,b,c,d,e)$ we have that all partial derivatives of $f$ are equal to $0$.

Thus, $$\frac{\partial f}{\partial a}=-\frac{1}{a^2}-\frac{33}{2(a+b+c+d+e)^2}+\lambda bcde=0$$ or
$$\lambda abcde=\frac{1}{a}+\frac{33a}{2(a+b+c+d+e)^2}$$
and since $\frac{\partial f}{\partial b}=0$ gives
$$\lambda abcde=\frac{1}{b}+\frac{33b}{2(a+b+c+d+e)^2},$$
we obtain:
$$(a-b)(2(a+b+c+d+e)^2-33ab)=0$$
and we can get a similar equality for all two variables in the minimum point,

which says that if for example $a\neq b$, $a\neq c$, $a\neq d$ and $a\neq e$ so $b=c=d=e$.

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 but very strong.

If $a=b$, but $a\neq c$, $a\neq d$ and $a\neq e$ so $c=d=e$.

Let $c=d=e=t^2$, where $t>0$.

Hence, $a=b=\frac{1}{t^3}$ and in this case we’ll get something obvious:
$$(t-1)^2(60t^8+120t^7+180t^6-9t^5-198t^4-92t^3+14t^2+120t+60)\geq0.$$

Done!

Replace with reciprocals so the problem becomes

$a + b + c + d + e + {33 \over 2}{1 \over {1 \over a} + {1 \over b} + {1 \over c} + {1 \over d} + {1 \over e}} \geq {83 \over 10}$

Sort the numbers so that $a\le b\le c\le d\le e$ and assume that $a,b,c,d,e$ are not all equal to $1$. Since $a<1$ and $e>1$ we have
$(a+b+c+d+e) – (1+b+c+d+ea) = a+e-1-ae = (e-1)(1-a) > 0$

Let $x_1=\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{e}$ and $x_2=\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}$

$x_1 – x_2 = \frac{1}{a}+\frac{1}{e}-\frac{1}{ea} – 1 = \frac{(e-1)(1-a)}{ea} >0$

The geometric harmonic inequality says $(bcdea)^{1/5}\ge \frac{5}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}} $ and thus using $abcde=1$ we conclude $\frac{1}{5}\ge \frac{1}{\frac{1}{1}+ \frac{1}{b}+ \frac{1}{c}+ \frac{1}{d}+ \frac{1}{ea}}$. Thus we have that $x_2\ge 5$.

The mean value theorem applied to the function $f(x)=1/x$ gives

$\frac{1}{x_1}-\frac{1}{x_2} = f'(\theta)(x_1-x_2)$ where $x_2 \le\theta\le x_1$. This tells us that since $\theta\ge x_2\ge 5$ that $f'(\theta)\ge f'(x_2) \ge f'(5) = -\frac{1}{5^2}$

Since $bcde=1/a$ we cannot have $b,c,d,e$ all less than $1/a^{1/4}$. As $e$ is the largest of them we must have $e\ge 1/a^{1/4}$ and since $0\le a\le 1$ we can conclude $ea\ge a^{3/4}\ge 1$.

Now putting all the computations above together gives
$\left(a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}}\right) – \left(1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\right) =$
$(e-1)(1-a) + \frac{33}{2}\left(\frac{1}{x_1}-\frac{1}{x_2}\right)= (e-1)(1-a) + \frac{33}{2}f'(\theta)(x_1-x_2) =$
$(e-1)(1-a) + \frac{33}{2}f'(\theta)\frac{(e-1)(1-a)}{ea}\ge (e-1)(1-a) – \frac{33}{50}\frac{(e-1)(1-a)}{ea}\ge $

$(e-1)(1-a) – \frac{33}{50}(e-1)(1-a) = \frac{27}{50}(e-1)(1-a)\ge 0$

Thus if we replace $a,b,c,d,e$ with $1,b,c,d,ea$ the left hand side of the inequality decreases while still maintaining $abcde=1$. We sort the numbers $1,b,c,d,ea$ and choose the smallest and largest among them and repeat the process we have just described. Eventually all five numbers will become 1 thus showing

$a+b+c+d+e+\frac{33}{2}\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}} \ge 1+b+c+d+ea+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{ea}}\ge \ldots \ge$

$1+1+1+1+1+\frac{33}{2}\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}+\frac{1}{1}}=\frac{83}{10}$