$-\lambda-\log(1-\lambda)\le\lambda^2$ on $[0,0.68]$

The elementary inequality behind B&V (9.50): for $0\le\lambda\le 0.68$,

Boyd & Vandenberghe state this after (9.49) and illustrate it in Figure 9.24 (p. 503): the solid curve $-(\lambda+\log(1-\lambda))$, which behaves like $\lambda^2/2$ for small $\lambda$, stays below the dashed parabola $\lambda^2$ on the interval $0\le\lambda\le 0.68$. Combining it with (9.49) yields the suboptimality bound $f(x)-p^\star\le\lambda(x)^2$ of (9.50).

The constant $0.68$ is essentially sharp: writing $h(\lambda)=\lambda^2+\lambda+\log(1-\lambda)$, one computes

so $h$ increases on $(0,\tfrac12)$, decreases on $(\tfrac12,1)$, and $h(0)=0$. Hence $h\ge 0$ on $[0,0.68]$ reduces to the single numerical check $h(0.68)\ge 0$, and indeed $h(0.68)=0.4624+0.68+\log(0.32)\approx 0.0030>0$, while already at $\lambda=0.7$ one has $h(0.7)\approx-0.0140<0$. So the claim is true but tight, and any proof must at some point bound $\log(0.32)$ numerically.

This is a self-contained one-variable real-analysis statement: it involves no convex optimization, and is separated out here so that the self-concordance argument of (9.49) and this numerical estimate can be proved independently.