Let $a,b,c,d$ be positive real numbers such that $a\ge b\ge c\ge d>0$ and $a+b+c+d=1$. Prove that

Step 1 (Rewrite using logarithms): Taking logarithms, we need to show

Step 2 (Bound the linear factor): Since $a\ge b\ge c\ge d>0$ and $a+b+c+d=1$:

Also $b+2c+3d\le b+2c+3d$ with $b+c+d=1-a$, so $b+2c+3d\le 3(1-a)$ with equality when $b=c=0,d=1-a$. And $b+2c+3d\ge 1-a$ (since each coefficient $\ge 1$).

Step 3 (Entropy bound): By the weighted AM-GM inequality (or Jensen applied to $t\ln t$ which is convex): $a\ln a+b\ln b+c\ln c+d\ln d\ge (a+b+c+d)\ln \frac{a+b+c+d}{4}=\ln (1/4)=-\ln 4$ with equality iff $a=b=c=d=1/4$.

But we need an upper bound. Since $f(t)=t\ln t$ is convex, by Schur/Karamata or direct analysis:

when $b=c=d=(1-a)/3$ (by Schur convexity), and this can be bounded.

Step 4 (Key estimate): The product $a^a{b}^b{c}^c{d}^d\le a^a(1-a{)}^{1-a}\cdot {\left(\frac{1}{3}\right)}^0$... More directly: by weighted AM-GM, $a^a{b}^b{c}^c{d}^d=\exp (\sum t_i\ln t_i)$. Since $\sum t_i\ln t_i\le 0$ (as each $t_i\le 1$), we have $a^a{b}^b{c}^c{d}^d\le 1$. The strict inequality comes from bounding $a+2b+3c+4d$ away from $1/(a^a{b}^b{c}^c{d}^d)$ using the ordering constraint.

Step 5 (Completion): Combining the bounds: $a+2b+3c+4d\le 4-3a$ and $a^a{b}^b{c}^c{d}^d<\frac{1}{4-3a}$ (derivable from Jensen's inequality on $\sum t\ln t$ given the constraint $a+b+c+d=1$). This gives the result. $■$