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Prove $(x+y)^{x+y}\ge x^x y^y$, subject to constraints $x>1$ and $y>1$. More generally, what other functions besides $x^x$ and $x\ln(x)$ satisfy this inequality? (I vaguely remember a paper where a well known distribution measure, which name unfortunately escapes me, was considered together with standard entropy).

Edit: For $x\ln(x)$ we are looking into $(x+y)\ln(x+y)\ge x\ln(x) + y\ln(y)$. In fact, here is original formulation of the problem: "Prove that the standard partition lattice order is consistent with total order of partition entropy values". The second part of my question is if there is alternative function on set partitions which agrees with lattice of partitions.

Edit2: I was looking into Gini Index: $\displaystyle\sum\limits_{i=0}^n p_i(1-p_i)$.

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Since $(x+y)^{x+y}=(x+y)^x(x+y)^y$ and $x,y$ are both positive, what can we conclude? –  Galois Group Jan 25 '12 at 23:40
@Fortuon Paendrag: Please make that an answer! –  André Nicolas Jan 25 '12 at 23:48
@AndréNicolas:Oh, ok. I was trying not to give away the answer so soon. But I will assume that enough time has elapsed. –  Galois Group Jan 25 '12 at 23:52
$x\ln(x)$ does not always satisfy the inequality $f(x+y) \ge f(x)f(y)$: for example $7 \ln(7) \lt 3\ln(3) \times 4\ln(4)$. –  Henry Jan 25 '12 at 23:52
$k^x$ turns the inequality into an equality. –  Henry Jan 25 '12 at 23:53

2 Answers 2

$(x+y)^{x+y}=(x+y)^x(x+y)^y \geq x^xy^y$, since $x+y \geq x,y \geq 0 $ and exponentiation is monotone.

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Take the logarithm of both sides. On the right, we get $x\log x+y\log y$. On the left we get $(x+y)\log(x+y)$, which is $x\log(x+y)+y\log(x+y)$. This is clearly bigger than the log of the right-hand side.

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