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

  • 3
    $\begingroup$ "Hoc" as in "ad hoc"? $\endgroup$
    – GEdgar
    Aug 12, 2011 at 14:36
  • $\begingroup$ Maybe he ment "How" $\endgroup$ Aug 12, 2011 at 14:40
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    $\begingroup$ -1: I think it would be nice if you started including some information on what you tried to do before asking the question here. Randomly checking among your 40 questions so far revealed not a single question that wasn't only a copy of a problem from some IMO-style list. $\endgroup$
    – t.b.
    Aug 12, 2011 at 21:44
  • $\begingroup$ Putting $a:=x$, $b:=x$, $c:=y$, $d:=y$ $e:=x^{-2}y^{-2}$ and plotting the left side of the stated inequality one obtains a nonconvex surface. This indicates that simple convexity arguments won't do. $\endgroup$ Dec 21, 2011 at 11:10

3 Answers 3


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?


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 $$297.5^2-4\cdot311\cdot75=297.5^2-311\cdot300<0.$$

  • $\begingroup$ Wait, $\mathbb{R}_{+}^5$ is not compact actually? As it is not bounded (even if you extend $f$ to $0$ via a limit). $\endgroup$ Nov 5, 2022 at 21:10
  • $\begingroup$ @DesmondMiles You are right. Thank you for your points! We can end the proof by the Lagrange multipliers method, of course. I have found an easier reasoning and posted a solution, for which we have one case only. See now, please. $\endgroup$ Nov 5, 2022 at 23:19

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

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


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    $\begingroup$ I don't see how $ea\ge a^{3/4} \ge 1$ follows; with $a\le 1$ this would require $a=1$. $\endgroup$
    – Zander
    Apr 6, 2013 at 13:19
  • $\begingroup$ Oh you're right that is a mistake. Unfortunately I don't see how to repair the problem. $\endgroup$
    – user782220
    Apr 10, 2013 at 9:51

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