= Solution
The increasing baseline order is $1/5,1/4,1/2,3/4$. A <water-filling algorithm> with three active coordinates gives
$$
\tau=\frac{1+1/5+1/4+1/2}{3}=\frac{13}{20},
\qquad \frac12<\frac{13}{20}<\frac34.
$$
Therefore, in the original coordinate order,
$$
\boxed{x^*=\left(\frac25,\frac3{20},\frac9{20},0\right).}
$$
The first three shifted coordinates all equal $13/20$, the fourth remains $3/4$, and the allocations sum to one. For an explicit <KKT> certificate take $\lambda=20/13$, $\nu_1=\nu_2=\nu_3=0$, and $\nu_4=20/13-4/3=8/39$. The <strict convexity> established above makes this optimum unique.
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