<s>[INST] 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:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} +\frac{1}{e}+ \frac{33}{2(a + b + c + d+e)} \ge{\frac{{83}}{10}}\ \ ?$$
 [/INST] 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?
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