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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    $\begingroup$ "Hoc" as in "ad hoc"? $\endgroup$ – GEdgar Aug 12 '11 at 14:36
  • $\begingroup$ Maybe he ment "How" $\endgroup$ – Beni Bogosel Aug 12 '11 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 '11 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. $\endgroup$ – Christian Blatter Dec 21 '11 at 11:10

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.


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

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


  • $\begingroup$ 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 '13 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 '13 at 9:51

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: $$(t-1)^2(60t^8+120t^7+180t^6-9t^5-198t^4-92t^3+14t^2+120t+60)\geq0.$$



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