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Can anyone offer some guidance on proving the following inequality? Define $\Lambda_1(a)=-a\log a$ and $\Lambda_2(a,b)=-(a+b)\log(a+b)$. Then if $a$, $b$, $c$, and $d$ are non-negative numbers summing to one, the following holds: \begin{align} \Lambda_2(a,b)+\Lambda_2(b,c)+\Lambda_2(c,d)+\Lambda_2(d,a)\geq \Lambda_1(a)+\Lambda_1(b)+\Lambda_1(c)+\Lambda_1(d). \end{align} I've tested a bunch of cases in Mathematica, so I'm pretty certain it's true. The concavity of $\Lambda_1$ gives an upper bound on the left-hand-side, so that doesn't seem to be the right approach. It's also straightforward to show that equality holds if $a=d$ and $b=c$, but I don't think that has much to do with the general case. I assume this would follow quickly from the right log inequality, so even just the name of such an inequality would be helpful.

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  • $\begingroup$ Are $a,b,c,d$ in $[0,1]$? $\endgroup$
    – Alex R.
    Jul 21 '15 at 0:53
  • $\begingroup$ Yes, they all are. $\endgroup$
    – Jeffrey
    Jul 21 '15 at 1:10
  • $\begingroup$ Started with $\log(a^2+b^2+c^2+d^2) \ge a \log(a) + b \log(b) + c \log(c) + d \log(d)$ but didn't get anywhere. Also perhaps simplify notation by $\Lambda_2(a,b)=\Lambda_1(a+b)$ $\endgroup$
    – karakfa
    Jul 21 '15 at 1:59
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By weighted AM/GM, \begin{align*} &\left(\frac{(a+b)(d+a)}{a}\right)^a \left(\frac{(b+c)(a+b)}{b}\right)^b \left(\frac{(c+d)(b+c)}{c}\right)^c \left(\frac{(d+a)(c+d)}{d}\right)^d \\ &\le a\cdot\frac{(a+b)(d+a)}{a} +b\cdot\frac{(b+c)(a+b)}{b} +c\cdot\frac{(c+d)(b+c)}{c} +d\cdot\frac{(d+a)(c+d)}{d} \\ &= (a+b)(d+a) +(b+c)(a+b) +(c+d)(b+c) +(d+a)(c+d) \\ &= \big((a+b)+(c+d)\big) \big((d+a)+(b+c)\big) \\ &= 1 \end{align*} Rearranging, $$ \frac1{a^a b^b c^c d^d} \le \frac1{(a+b)^{a+b} (b+c)^{b+c} (c+d)^{c+d} (d+a)^{d+a}} $$ Taking logs yields the desired inequality.

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  • $\begingroup$ Very nice application +1! $\endgroup$
    – Macavity
    Jul 21 '15 at 3:31
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Since $a,b,c,d$ are nonnegative and sum to 1, consider them as probabilities in some sample space with outcomes $A,B,C,D$ respectively having probabilities $a,b,c,d$.

Without loss of generality, let $U$ be the event $A\cup B$, and $V$ be the event $B\cup C$. Let $H(X)=\sum_{i=1}^m -p_i \log p_i$ be the standard entropy functional for a discrete random variable $X$ with p.m.f. $\{p_1,p_2,...p_m\}$. Then, the LHS of the original problem is $H(\mathbf{1}_U)+H(\mathbf{1}_V)$ while the RHS is $H(\mathbf{1}_U,\mathbf{1}_V)$ according to standard notation.

From elementary information theory (via Jensen's inequality primarily) we have $H(\mathbf{1}_U)+H(\mathbf{1}_V) = H(\mathbf{1}_U,\mathbf{1}_V) + I(\mathbf{1}_U;\mathbf{1}_V)$ where $I(\mathbf{1}_U;\mathbf{1}_V)\ge 0$ is the mutual information between $\mathbf{1}_U$ and $\mathbf{1}_V$. The inequality in the original question follows.

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Just a trial...

From the definitions follow

$$ \Lambda_2(a,b) = \Lambda_1(a+b). $$

So need to show that

$$ \log\big( a^a b^b c^c d^d \big) - \log\big( (a+b)^{a+b} (b+c)^{b+c}(c+d)^{c+d}(d+a)^{d+a} \big) \ge 0. $$

First we rewrite this as $$ (abcd) \log\big( abcd \big) - (a+b)(b+c)(c+d)(d+a) \log\big( (a+b)(b+c)(c+d)(d+a) \big) \ge 0. $$

Let us write $$ p = abcd, \hspace{2em} q = (a+b)(b+c)(c+d)(d+a). $$

We have $$ q = (a+b)(b+c)(c+d)(d+a) < (a+b+c+d)^4 = p^4, $$ but also $$ p < q. $$

Due to the summation to one, we have $$ a+b+c+d = 1 \Rightarrow p = abcd \le 4^{-4}. $$

Note that $x \log(x)$ is decreasing for $0 \le x < e^{-1}$.

So clearly we have $$ 0 < p < q < 4^{-1} < e^{-1}. $$

And consequently we obtain $$ p \log\big(p\big) \ge q \log\big( q \big). $$

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