最大拉普拉斯能量猜想的一个证明：通过一个尖锐的特征值求和界
A proof of the maximum Laplacian energy conjecture for connected graphs via a sharp eigenvalue-sum bound
中文总结 AI 辅助
通过证明一个尖锐的拉普拉斯特征值求和界，解决了关于连通图最大拉普拉斯能量的猜想，并确定了唯一极值图。
中文摘要 AI 辅助
设$S_k(G)$表示$n$阶$m$边连通图$G$的$k$个最大拉普拉斯特征值之和。记$\mathrm{PA}_{n,\omega}$为从一个$\omega$顶点团的一个顶点上连接$n-\omega$个悬挂顶点所得的图，并设\\[ M_{n,k}:=\binom{k+1}{2}+n-k-1,\\] 即$\mathrm{PA}_{n,k+1}$的边数。对于$n/2<k\le n-2$，我们证明尖锐界\\[ S_k(G)\le \frac{2k}{n}m+ \frac{2(n-k)}{n}M_{n,k}-(n-k-1),\\] 等号由$\mathrm{PA}_{n,k+1}$取得。该界与Brouwer不等式互补，并在$m<M_{n,k}$时严格更强。结合我们的界与Brouwer不等式，我们解决并加强了Vinagre、Del-Vecchio、Justo和Trevisan的一个猜想：对每个$n$，菠萝图$\mathrm{PA}_{n,1+\lfloor2n/3\rfloor}$在所有$n$阶连通图中最大化拉普拉斯能量；此外，当$n>4$时，它是唯一的最大化图。
英文摘要
Let $S_k(G)$ denote the sum of the $k$ largest Laplacian eigenvalues of a connected graph $G$ of order $n$ and size $m$. Write $\mathrm{PA}_{n,ω}$ for the graph obtained from an $ω$-vertex clique by attaching $n-ω$ pendant vertices to one of its vertices, and set \[ M_{n,k}:=\binom{k+1}{2}+n-k-1, \] the number of edges of $\mathrm{PA}_{n,k+1}$. For $n/2<k\le n-2$, we prove the sharp bound \[ S_k(G)\le \frac{2k}{n}m+ \frac{2(n-k)}{n}M_{n,k}-(n-k-1), \] with equality attained by $\mathrm{PA}_{n,k+1}$. This bound is complementary to Brouwer's inequality and is strictly stronger when $m<M_{n,k}$. Combining our bound with Brouwer's inequality, we resolve and strengthen a conjecture of Vinagre, Del-Vecchio, Justo, and Trevisan: for every $n$, the pineapple $\mathrm{PA}_{n,1+\lfloor2n/3\rfloor}$ maximizes the Laplacian energy among all connected graphs of order $n$; moreover, for $n>4$, it is the unique maximizer.
