logarithms - 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.

Answer

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.