a log inequality 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.
 A: 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.
A: 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.
A: 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
$$
p \log\big(p\big) \ge q \log\big( q \big).
$$
