TITLE: Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$
QUESTION [7 upvotes]: Let $a,b,c,d,e$ be positive real numbers which satisfy $abcde=1$. How can one prove that:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$

REPLY [2 votes]: 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!