# Some New Bounds for the Inverse Sum Indeg Energy of Graphs
## 4 Bounds for the ISI Energy of Graphs  
For convenience, we let \(\gamma_{1}\geq\gamma_{2}\geq\ldots\geq\gamma_{n}\) as the absolute values of eigenvalues \(\mu_{i},1\leq i\leq n\), which are arranged in decreasing order. It is obvious that  
\[\epsilon_{isi}(G)=\sum_{i=1}^{n}|n_{i}|=\sum_{i=1}^{n}\gamma_{i}\]  
and  
\[tr(\mathcal{C}^{2})=\sum_{i=1}^{n}\mu_{i}^{2}=\sum_{i=1}^{n}\gamma_{i}^{2}\.\]  
Theorem 4.1: _Let \(G\) be a graph of order \(n\) and size \(m\), with minimum degree \(\delta\) and maximum degree \(\Delta\). Then_  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\ . \tag{22}\]  
_The equality holds if and only if \(G\cong\overline{K_{n}}\) or \(G\cong\frac{n}{2}K_{2}\)._  
Proof: Bear in mind that \(\gamma_{1}^{2},\gamma_{2}^{2},\ldots,\gamma_{n}^{2}\) form the eigenvalues of \(\mathcal{C}^{2}\). Combined this fact with Lemma 2 we obtain  
\[\epsilon_{isi}(G)=\sum_{i=1}^{n}\gamma_{i}\leq\sqrt{\sum_{i=1}^{n}1}\sqrt{ \sum_{i=1}^{n}\gamma_{i}^{2}}=\sqrt{ntr(\mathcal{C}^{2})}. \tag{23}\]  
Lemma 13 implies that  
\[tr(\mathcal{C}^{2})=2\sum_{\gamma_{i}\gamma_{j}\in\mathcal{E}(G)}\frac{(d_{i} d_{j})^{2}}{(d_{i}+d_{j})^{2}}\leq\frac{m\Delta^{2}}{2}.\]  
Hence, we have  
\[\epsilon_{isi}(G)\leq\frac{\Delta}{2}\sqrt{2 mn}\.\]  
The equality in (23) is clearly attained when \(G\cong\overline{K_{n}}\) or \(G\cong\frac{n}{2}K_{2}\). If the equality in (22) is achived, then equality must hold in (23). So we have that \(\gamma_{1}=\gamma_{2}=\ldots=\gamma_{n}\). Hence, \(G\cong\frac{n}{2}K_{2}\) or \(G\cong\overline{K_{n}}\). Theorem 4.2: _Let \(G\) be an \(n\)-vertex graph, we have_  
\[\epsilon_{isi}(G)\geq(n-1)\gamma_{1}+\sqrt{tr(\mathcal{C}^{2})-(n-1)\gamma_{1} ^{2}}\, \tag{24}\]  
_and the equality is obtained iff \(G\cong\overline{K_{n}}\) or \(G\cong\frac{n}{2}K_{2}\)._  
Proof: Setting \(w_{i}=\gamma_{i}\), \((i=1,2,\ldots,n)\), inequality (8) becomes  
\[\frac{1}{n}\sum_{i=1}^{n}\gamma_{i}+\sqrt{\frac{1}{n(n-1)}\sum_{i=1}^{n}\! \left(\gamma_{i}-\frac{1}{n}\sum_{i=1}^{n}\gamma_{i}\right)^{2}}\]  
\[=\frac{\epsilon_{isi}(G)}{n}+\sqrt{\frac{1}{n(n-1)}\sum_{i=1}^{n}\!\left( \gamma_{i}-\frac{\epsilon_{isi}(G)}{n}\right)^{2}}\]  
\[=\frac{\epsilon_{isi}(G)}{n}+\sqrt{\frac{1}{n(n-1)}\!\left(\sum_{i=1}^{n}\gamma _{i}^{2}-\frac{2\epsilon_{isi}(G)}{n}\sum_{i=1}^{n}\gamma_{i}+\sum_{i=1}^{n} \frac{(\epsilon_{isi}(G))^{2}}{n^{2}}\right)}\]\[tr(\mathcal{C}^{2})=\sum_{i=1}^{n}\mu_{i}^{2}=\sum_{i=1}^{n}\gamma_{i}^{2}\.\]0  
Inequality (23) implies that \(tr(\mathcal{C}^{2})\geq\frac{(\varepsilon_{isi}(G))^{2}}{n}\). Hence,  
\[tr(\mathcal{C}^{2})=\sum_{i=1}^{n}\mu_{i}^{2}=\sum_{i=1}^{n}\gamma_{i}^{2}\.\]1  
Assume that equality in (24) is achieved.Then \(\gamma_{i}^{2}=\frac{1}{n}\sum_{i=1}^{n}\gamma_{i}^{2}\) for any \(1\leq i\leq n\). So, we have \(\gamma_{1}=\gamma_{2}=\ldots=\gamma_{n}\). By Lemma 13, we know that \(G\cong K_{n}\) or \(G\cong\frac{n}{4}K_{2}\). Conversely, the equality in (24) holds obviously for \(G\cong\frac{n}{2}K_{2}\) or \(G\cong\overline{K_{n}}\). This completes the proof. As a generalisation of the Shisha-Mond inequality, the Klamkin-McLenaghan inequality can be stated as follows. **Lemma 20** ((Klamkin-McLenaghan inequality) [58]). : _Let \(X=(x_{1},x_{2},\ldots,x_{n})\) and \(Y=(y_{1},y_{2},\ldots,y_{n})\) be \(n\)-tuples of non-negative real numbers satisfying \(0\leq m\leq\frac{x_{i}}{y_{i}}\leq M\) for each \(i\in\{1,2,\ldots,n\}\), and \(w_{i}\geq 0\). Then,_  
\[tr(\mathcal{C}^{2})=\sum_{i=1}^{n}\mu_{i}^{2}=\sum_{i=1}^{n}\gamma_{i}^{2}\.\]2  
**Theorem 7**. : _Let \(G\) be a graph of order \(n\). Then, the following inequality is valid:_  
\[tr(\mathcal{C}^{2})=\sum_{i=1}^{n}\mu_{i}^{2}=\sum_{i=1}^{n}\gamma_{i}^{2}\.\]3  
_Equality is attained if and only if \(G\cong\overline{K_{n}}\) or \(G\cong\frac{n}{2}K_{2}\)._  
Proof. : Setting \(x_{i}=\gamma_{i}\), \(y_{i}=1\), and \(w_{i}=1\), \((i=1,2,\ldots,n)\), then \(0\leq\gamma_{n}\leq\frac{x_{i}}{y_{i}}\leq\gamma_{1}\) for each \(i\in\{1,2,\ldots,n\}\).