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Can anyone show, step-by-step, how the expression on the LHS can be turned into the expression on the RHS?

$x^ay^b=a^ab^b(a+b)^{-(a+b)}(x+y)^{a+b}$

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    $\begingroup$ Seems to be false identity, could you recheck? $\endgroup$ – Juanito Jul 1 '14 at 22:25
  • $\begingroup$ could you rewrite the expression? it doesn't seem right. $\endgroup$ – Vishwa Iyer Jul 1 '14 at 22:25
  • $\begingroup$ I doubt it can be done. Are you sure that you wrote the problem completely and without typos? This seems to be some sort of equation, not an identity. $\endgroup$ – TZakrevskiy Jul 1 '14 at 22:26
  • $\begingroup$ this is exactly as it appears in the textbook! so either there's a mistake in the book, or it's just super complicated. $\endgroup$ – Gauss3000 Jul 1 '14 at 22:27
  • $\begingroup$ I didn't know how to type in the "identity" symbol here instead of the "equals" symbol, but the textbook gives this as an identity. How close can you guys get the LHS to something that vaguely resembles the RHS? $\endgroup$ – Gauss3000 Jul 1 '14 at 22:28
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Assuming all the numbers are positive, we in fact have $$ x^a y^b \le a^a b^b (a+b)^{-(a+b)} (x+y)^{a+b} \tag{$\ast$} $$ with equality if and only if $x/y = a/b$.

This is a standard application of the AM/GM inequality, as follows: \begin{align*} &\left(\frac{x}{x+y}\cdot\frac{a+b}{a}\right)^{a/(a+b)} \left(\frac{y}{x+y}\cdot\frac{a+b}{b}\right)^{b/(a+b)} \\ &\le \frac{a}{a+b}\left(\frac{x}{x+y}\cdot\frac{a+b}{a}\right) + \frac{b}{a+b} \left(\frac{y}{x+y}\cdot\frac{a+b}{b}\right) \\ &= 1 \end{align*} Raising both sides to the power $a+b$ and rearranging yields ($\ast$), with equality iff $$ \frac{x}{x+y}\cdot\frac{a+b}{a} = \frac{y}{x+y}\cdot\frac{a+b}{b} $$ which is equivalent to $\frac xy = \frac ab$.

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Take x=0, y=1, a=1, b=1. The LHS is not equal to RHS. So, the identity is False.

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