# Some New Bounds for the Inverse Sum Indeg Energy of Graphs
## 4 Bounds for the ISI Energy of Graphs
Theorem 12: _Let \(G\) be a connected graph of order \(n>1\) with \(m\) edges and minimum degree \(\delta\). Then,_  
\[tr(\mathcal{C}^{2})=2\sum_{\gamma_{i}\gamma_{j}\in\mathcal{E}(G)}\frac{(d_{i} d_{j})^{2}}{(d_{i}+d_{j})^{2}}\leq\frac{m\Delta^{2}}{2}.\]7  
_and the equality holds if and only if \(G\cong K_{\frac{n}{2},\frac{n}{2}}\)._  
Proof: For \(n=2\), \(G=K_{1,1}\) and, hence, the equality holds. Otherwise, \(n\geq 3\). From Lemma 14, we know that the sum of the eigenvalues of \(\mathcal{C}\) is zero, and we can deduce that  
\[tr(\mathcal{C}^{2})=2\sum_{\gamma_{i}\gamma_{j}\in\mathcal{E}(G)}\frac{(d_{i} d_{j})^{2}}{(d_{i}+d_{j})^{2}}\leq\frac{m\Delta^{2}}{2}.\]8  
Therefore, we have  
\[tr(\mathcal{C}^{2})=2\sum_{\gamma_{i}\gamma_{j}\in\mathcal{E}(G)}\frac{(d_{i} d_{j})^{2}}{(d_{i}+d_{j})^{2}}\leq\frac{m\Delta^{2}}{2}.\]9Combining the definition of ISI energy and Lemmas 1 and 14, we obtain  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]0 \[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]1 \[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]2 \[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]3  
and inequality (32) follows. This concludes the first part of the proof. Suppose now that the equality holds in (32). Then, all the above inequalities must be equalities. Equality in (34) implies that \(\frac{d_{i}^{2}d_{j}^{2}}{(d_{i}+d_{j})^{2}}=\frac{\delta^{2}}{4}\) for each edge \(v_{i}v_{j}\in E(G)\); that is, \(d_{i}=d_{j}\) for each edge \(v_{i}v_{j}\in E(G)\). As \(G\) is assumed to be connected, it is regular. From equality in (33), we see that there are two nonzero eigenvalues and all the remaining eigenvalues are zero; that is, \(\mu_{1}=-\mu_{n}\) and \(\mu_{i}=0\) for \(2\leq i\leq n-1\). Since \(G\) is regular, \(\frac{\delta}{2}\lambda_{i}=\mu_{i}\) for all \(1\leq i\leq n\). Therefore, \(\lambda_{1}=-\lambda_{n}\) and \(\lambda_{i}=0\) for \(2\leq i\leq n-1\). Since \(G\) is connected, by Lemma 18, it must be \(G\cong K_{\frac{n}{2},\frac{n}{2}}\). Conversely, by direct checking, we verify that equality holds in (31) for \(G\cong K_{\frac{n}{2},\frac{n}{2}}\). This completes the proof. Theorem 4.1: _Let \(G\) be a graph of order \(n>1\) and size \(m\). Then,_  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]4  
_where \(G^{\prime}\) is the graph obtained from \(G\) by deleting all isolated vertices. The equality holds if and only if \(G\cong\overline{K_{n}}\) or \(G\cong K_{p,p}\cup(n-2p)K_{1}\) and \(p=1,2,\ldots,\lfloor\frac{n}{2}\rfloor\)._  
Proof: For \(m=0\), we have \(G=\overline{K_{n}}\) and, hence, the equality holds. Otherwise, \(m\geq 1\). Let \(p\) be the number of isolated vertices and let \(k\) be the number of connected components in \(G\). In addition, let \(G_{i}\) be the \(i\)-th connected component of \(G\) with order \(n_{i}\geq 2\), \(m_{i}\geq 1\) edges and minimum degree \(\delta_{i}\). Hence, we have \(n=p+\sum_{i=1}^{k}n_{i}\), \(m=\sum_{i=1}^{k}m_{i}\) and \(\delta_{i}\geq\delta(G^{\prime})\). Without loss of generality, we may assume that \(m_{1}\geq m_{2}\geq\ldots\geq m_{k}\geq 1\). By Theorem 4.1, we have  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]5  
Notice that, for positive real numbers \(a\) and \(b\), \((a\geq b)\),  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]6  
with equality if and only if \(b=0\). Applying this result to (36) and by Lemma 19, we obtain  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]7 \[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]8 \[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]9 \[\epsilon_{isi}(G)\geq(n-1)\gamma_{1}+\sqrt{tr(\mathcal{C}^{2})-(n-1)\gamma_{1} ^{2}}\, \tag{24}\]0\[\epsilon_{isi}(G)\geq(n-1)\gamma_{1}+\sqrt{tr(\mathcal{C}^{2})-(n-1)\gamma_{1} ^{2}}\, \tag{24}\]1 This concludes the first part of the proof. Suppose that the equality holds in (35) for \(m\geq 1\). Then, all the above inequalities must be equalities. Since \(m_{i}\geq 1\), \(1\leq i\leq k\), we must have \(k=1\). By Theorem 12, we then have \(G_{1}\cong K_{\frac{n_{1}}{2},\frac{n_{1}}{2}}\).