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Let $u,v,w>0$ and $a,b,c$ are positive constant. Prove that $\left(\frac{u}{a}\right)^a.\left(\frac{v}{b}\right)^b.\left(\frac{w}{c}\right)^c \le \left(\frac{u+v+w}{a+b+c}\right)^{(a+b+c)} $

First, I prove with $x+y+z=1$ so $x^ay^bz^c\le\left(\frac{a}{a+b+c}\right)^a\left(\frac{b}{a+b+c}\right)^b\left(\frac{c}{a+b+c}\right)^c$ by Largrange theorem

And it become $\left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c}=\left(\frac{x+y+z}{a+b+c}\right)^{a+b+c}$

So it true with $x+y+z=1$ but i can't prove it true with $x,y,z>0$. Please help me! Thank you.

tompi2394
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    I am sure that this solution https://math.stackexchange.com/q/593749/42969 can be generalized from two to three variables. – Martin R Aug 28 '21 at 10:34
  • @MartinR Sorry but in this case I don't think the question deserves closure as a duplicate. Of course the reference you have given is very useful (and they worth to be linked) but this is a well distinct question. Moreover this question is well posed showing the effort by the asker. Can we say the same for the alleged duplicate? – user Aug 28 '21 at 10:48
  • @user: It is exactly the same for two, three, or 77 variables: Take the logarithm and use Jensen's inequality. – Martin R Aug 28 '21 at 10:50
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    @MartinR I understand your point but this is precisely the case of a good duplicate. I think that an answer for this one would enrich the site, maybe also with some other different method. if we close suddenly the question, how can we expect this new contributions can be made? For the quality of the questions and in the spirit of a generalization, maybe it should be better to have an answer here and close the previous one, if we want close one question. – user Aug 28 '21 at 10:54
  • @MartinR Oh thank you, I don't know this problem can solve by Jensen. So i have another solve, but i want to know why my solve wrong so please check my solve. Thank you very much – tompi2394 Aug 28 '21 at 12:34

2 Answers2

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The inequality is homogenenous in $(u, v, w)$: If you have proven $$ \left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c\le\left(\frac{1}{a+b+c}\right)^{a+b+c} $$ under the condition $x+y+z=1$ then you can use that to prove the general case: With $$ x = \frac{u}{u+v+w}, y = \frac{v}{u+v+w},z = \frac{w}{u+v+w} $$ one has $x+y+z=1$ and $$ \left(\frac{u}{a}\right)^a\left(\frac{v}{b}\right)^b\left(\frac{w}{c}\right)^c = \left(\frac{x}{a}\right)^a\left(\frac{y}{b}\right)^b\left(\frac{z}{c}\right)^c (u+v+w)^{a+b+c} \\ \le \left(\frac{1}{a+b+c}\right)^{a+b+c}(u+v+w)^{a+b+c} = \left(\frac{u+v+w}{a+b+c}\right)^{a+b+c} \, . $$

Martin R
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Since we have positive numbers involved therefore we can use weighted $A.M-G.M$ on $\frac{u}{a},\frac{v}{b},\frac{w}{c}$ with weights respectively $a,b,c$

$$\displaystyle\bigg({\frac{\sum_ia_ix_i}{\sum_i a_i}}\bigg)^{\sum_i a_i}\geq\bigg(\Pi (x_i)^{(a_i)}\bigg) $$

Here $x_1=\frac{u}{a} $ and $a_1=a$ , $x_2=\frac{v}{b} $ and $a_2=b$ and $x_3=\frac{w}{c} $ and $a_3=c$

Equality holds when $\frac{u}{a}=\frac{v}{b}=\frac{w}{c}$

Lalit Tolani
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