Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
without $x+y+z=1$, the equality is OK also. –  chenbai Nov 4 '13 at 7:15
    
$ze^a+xe^b+ye^c\ge e^{\frac{az+bx+yc}{x+y+z}}=1$ –  chenbai Nov 4 '13 at 7:22
    
why without this problem is also? –  user94270 Nov 4 '13 at 8:02

3 Answers 3

up vote 7 down vote accepted
+50

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.

share|improve this answer
    
this is the right answer.+1 –  chenbai Nov 10 '13 at 5:54
    
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! –  Ron Ford Nov 12 '13 at 10:25
2  
The second solution is wrong. $b\log(a)<a\log(a)$ because $\log(a)<0$. –  Hansen Dec 21 '13 at 21:29
1  
@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

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.

share|improve this answer
    
+1. Thank you for correction me! –  Ron Ford Dec 28 '13 at 13:41
    
No problem. I like your solution. –  Hansen Dec 28 '13 at 16:55

Hint :

Use AM-GM inequality on all three terms and it will reduce to

$(x+y+z)\geqslant 3(xyz)^{1/3}$ which is true if you considered AM GM inequality on x,y,z

I worked out the problem with AM - GM Inequality and it works

Thanks

Satish

I made a mistake in the exponents and carried it through to the other terms and hence the problem. I understand that it is a mistake.

But, I tried something else and could not finish it through.

Original expression on the Left hand side

x^y/y^x + y^z/z^y + z^x/x^z ....... (1)

Try x=e^y, y=e^z, z=e^x

Now the transformed left hand side would be

e^y^2/e^zx + e^z^2/e^xy + e^x^2/e^yx ......(2)

Try AM GM Inequality on the transformed Left hand side

LHS >= 3* e^(x^2+y^2+z^2-zx-xy-yz)

and further reduce RHS

After this transformation, the original expression has the same value as the transformed expression and I think RHS could be reduced using AM GM inequality but I could not proceed further. I input both the LHS expressions (original and transformed) and get the same value which means the transformation is right. It is just that I cannot find any tool to further reduce it.

share|improve this answer
    
This is wrong.But Thank you –  user94270 Nov 3 '13 at 15:41
    
Please look at my edit and see if you could take it further –  satish ramanathan Nov 3 '13 at 21:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.