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let $x,y,z$ be positive numbers, and such $x+y+z=1$

show that $$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$

My try:

let $$a=\ln{\dfrac{x^y}{y^x}},b=\ln{\dfrac{y^z}{z^y}},c=\ln{\dfrac{z^x}{x^z}}$$ so $$a=y\ln{x}-x\ln{y},b=z\ln{y}-y\ln{z},c=x\ln{z}-z\ln{x}$$ and we note $$az+bx+yc=(y\ln{x}-x\ln{y})z+(z\ln{y}-y\ln{z})x+(x\ln{z}-z\ln{x})y=0$$ so $$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}=e^a+e^b+e^c$$ so $$\Longleftrightarrow e^a+e^b+e^c\ge 3$$ But then I can't prove it.

If this problem is to prove $$ze^a+xe^b+ye^c\ge 3,$$ I can prove it,because $$ze^a+xe^b+ye^c\ge=\dfrac{z}{x+y+z}e^a+\dfrac{x}{x+y+z}e^b+\dfrac{y}{x+y+z}e^c$$ so use Jensen's inequality,we have $$ze^a+xe^b+ye^c\ge e^{\dfrac{az+bx+yc}{x+y+z}}=3$$

This problem comes from How prove this $\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$?

Thank you very much!

Community
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3 Answers3

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Another solution. Without loss of generality, we can assume that $0<x\le y\le z$ and so $x=az$, $y=bz$, $0<a\le b\le1$. By the AM-GM inequality, we find that \begin{align} \frac{x^y}{y^x}+\frac{y^z}{z^y}+\frac{z^x}{x^z}&\ge 3\sqrt[3]{x^{y-z}y^{z-x}z^{x-y}}=3\sqrt[3]{(az)^{(b-1)z}(bz)^{(1-a)z}z^{(a-b)z}}=3\sqrt[3]{a^{(b-1)z}b^{(1-a)z}}=\\ &=3(a^{b-1}b^{1-a})^{z/3} \end{align} so we only need to prove that $a^{b-1}b^{1-a}\ge1$, which Hansen did in the answer below. We also see that the condition $x+y+z=1$ is not needed.

Kortlek
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  • this is the right answer.+1 – chenbai Nov 10 '13 at 05:54
  • Great answer, Walt! Do you have a bitcoin wallet? – Anonymous - a group Nov 12 '13 at 10:13
  • @Anonomous No, but I have a reputation wallet! – Kortlek Nov 12 '13 at 10:25
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    The second solution is wrong. $b\log(a)<a\log(a)$ because $\log(a)<0$. – Hans Dec 21 '13 at 21:29
  • However, the solution can be salvaged by looking at $g(x) = -(1-x)\log(a)+(1-a)\log(x)$. It is easy to see $g$ is a concave function with $g(a)=g(1)=0$. So, $g(b)>0,,\forall a<b<1$. – Hans Dec 21 '13 at 21:45
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    @Hansen, I see you tried to edit Karl Marx's post to add your correction. Please, don't do it; instead, you can add another answer with the correction: "... this answer is a correction for..." – Ian Mateus Dec 27 '13 at 20:49
  • @IanMateus: I did this because I did not want to claim credit for the solution as the main idea was Karl Marx'. But if writing a new answer is the protocol, I will do so. – Hans Dec 27 '13 at 20:54
  • @Hansen It's been a while since you warned him, you surely can do it :-) – Ian Mateus Dec 27 '13 at 21:00
  • The Hint part is wrong. The minimum of the right hand side is $1$ not $3$. – Hans Apr 13 '14 at 18:04
  • Please note the objection from @yegulalp below. It unfortunately invalidates the proof premised on the invariance over variable permutation. Do you have any idea mending the proof or for a new one? – Hans Apr 16 '15 at 16:04
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This is to prove $a^{b-1}b^{1-a}\ge 1$ as needed in Ron Ford's answer above. Let $f(b)=(1-b)(-\log a)-(1-a)(-\log b)$, $b\in [a,1]$. $f(a)=f(1)=0$. $f$ is concave, as $f''(b)=-\frac{1-a}{b^2}<0$. So $f(b)>0,\,\forall b\in(a,1)$ and the result follows.

Hans
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The inequality as posed in the OP appears to be correct. The accepted proof (Ford/Hansen) is not.

Note that the proof doesn't require the assumption that $x+y+z=1$. Take $x=6$, $y=2$, and $z=1$. Then the left side is ${36\over 64} + 2 + {1\over 6} = 2 + {35\over 48} < 3$, violating the inequality. Clearly the proof cannot be correct as stated.

The error in the Ford/Hansen proof is the assumption we can take $x \leq y \leq z$. This is not generally possible because the LHS of the inequality is not invariant to permuting any pair of the variables. Note that if we swap $x$ with $y$, the first term becomes its own reciprocal and the second and third terms become each other's reciprocals:

Define $$A = x^{y}/y^{x}, B = y^{z}/z^{y}, C = z^{x}/x^{z}.$$ Exchanging $x$ with $y$ makes $A \rightarrow 1/A$, $B \rightarrow 1/C$, and $C \rightarrow 1/B$.

Furthermore, no proof using the AM-GM inequality can work. Applying AM-GM we get: $$ A + B + C \geq 3(ABC)^{1/3}.$$ If $ABC < 1$, we can't finish the proof. If $ABC > 1$, then consider the case where we swap $x$ and $y$: $$ A^{-1} + B^{-1} + C^{-1} \geq 3(ABC)^{-1/3}$$ Unless $ABC=1$, the bound is insufficient in one of the two cases above to prove the desired result. If the bound works for a given $x$, $y$, and $z$ then it won't suffice for the case where we swap $x$ and $y$.

yegulalp
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  • I've removed the erroneous statements and numerical example to which Daniel responded. – yegulalp Aug 06 '14 at 12:51
  • Your objection is correct. Any ideas for a proof or counterexample? Is the reason you say this inequality "appears to be correct" some numerical verification? – Hans Apr 16 '15 at 15:51