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Known facts:

Let $a_i$, $b_i$, $i=1, \ldots, n$ be positive real numbers such that $a_1+ \cdots + a_n = a_1b_1 + \cdots +a_nb_n = 1$. Then $$b_1^{a_1}b_2^{a_2} \cdots b_n^{a_n} \leq 1.$$


Let $c_i$, $i=1, \ldots, n$ be positive real numbers such that $c_1c_2 \cdots c_n = 1$. Then $$c_1^{c_1} c_2^{c_2} \cdots c_n^{c_n} \geq 1.$$

Questions: Let $x_1, x_2, \cdots, x_n$ be positive real numbers. Prove that $$x_1^{x_1} x_2^{x_2} \cdots x_n^{x_n} \leq \left( \frac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots+x_n}\right)^{x_1+x_2+\cdots+x_n}.$$

It is a homework question. I would appreciate any help but not giving me the complete answers. Here comes my thoughts. In order to make use of the two facts, it is essential for me to construct the corresponding $a_i$, $b_i$ and $c_i$. This really bothers me. All I can think of is $$a_i = \frac{x_i}{x_1+\cdots+x_n}.$$ How should I proceed to find $b_i$ and $c_i$? Thanks in advance.

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    $\begingroup$ Hint:Do you enough familier with waighted Arithmatic or geometric mean? Use waighted AM-GM Inequality. $\endgroup$
    – user249332
    Commented Aug 28, 2015 at 16:11

2 Answers 2

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$S= \frac{x_1^2+x_2^2+\cdots+x_n^2}{x_1+x_2+\cdots+x_n}$

Try: $a_i = \frac{x_i}{x_1+\cdots+x_n}$ and $b_i = \frac{x_i}{S}$

Update

It works for sure, I just prove it.

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  • $\begingroup$ It really works. Thx! But how did you come up with the $a_i$ and $b_i$ so quickly? $\endgroup$
    – Nighty
    Commented Aug 28, 2015 at 16:35
  • $\begingroup$ @LeeKM I'm very smart :) $\endgroup$
    – user261263
    Commented Aug 28, 2015 at 16:36
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Assign weights $x_i$ on $x_i$'s , i.e. the weight of $x_1$ is $x_1$; that of $x_2$ is $x_2$ and so on.

So then, the weighted AM of $(x_1, x_2, \dots, x_n)=\frac {{x_1}^2+{x_2}^2\dots+{x_n}^2}{x_1+x_2+\dots+x_n}$ and the weighted GM is ${({x _1}^{x_1}{x_2}^{x_2}\dots {x_n}^{x_n})^{\frac 1{x_1+x_2+\dots+x_n}}}$

As, weighted AM$\ge$weighted GM, hence, your claim follows.

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