# Energy of Pythagorean Fuzzy Graphs with Applications
## 2 Energy of Pythagorean Fuzzy Graphs
If \(n\leq 2\sum\limits_{1\leq i<j\leq n}(\mu_{\mathcal{Q}}(z_{i}z_{j}))^{2}\) and \(n\leq 2\sum\limits_{1\leq i<j\leq n}(\
u_{\mathcal{Q}}(z_{i}z_{j}))^{2}\), then:_  
**(i)**: \(E(\mu_{\mathcal{Q}}(z_{i}z_{j}))\leq\frac{2\sum\limits_{1\leq i<j\leq n}(\mu_ {\mathcal{Q}}(z_{i}z_{j}))^{2}}{n}+\sqrt{(n-1)\left\{2\sum\limits_{1\leq i<j \leq n}(\mu_{\mathcal{Q}}(z_{i}z_{j}))^{2}-\left(\frac{2\sum\limits_{1\leq i<j \leq n}(\mu_{\mathcal{Q}}(z_{i}z_{j}))^{2}}{n}\right)^{2}\right\}}\)_;_
**(ii)**: \(E(\
u_{\mathcal{Q}}(z_{i}z_{j}))\leq\frac{2\sum\limits_{1\leq i<j\leq n}(\
u_ {\mathcal{Q}}(z_{i}z_{j}))^{2}}{n}+\sqrt{(n-1)\left\{2\sum\limits_{1\leq i<j \leq n}(\
u_{\mathcal{Q}}(z_{i}z_{j}))^{2}-\left(\frac{2\sum\limits_{1\leq i<j \leq n}(\
u_{\mathcal{Q}}(z_{i}z_{j}))^{2}}{n}\right)^{2}\right\}}\)_._  
**Proof. ** If \(A=[a_{ij}]_{n\times n}\) is a symmetric matrix with zero trace, then \(\lambda_{\max}\geq\frac{2\sum\limits_{1\leq i<j\leq n}a_{ij}}{n}\), where, \(\lambda_{\max}\) is the maximum eigenvalue of \(A\). If \(A(\mathcal{G})\) is the adjacency matrix of a PFG \(\mathcal{G}\), then \(\lambda_{1}\geq\frac{2\sum\limits_{1\leq i<j\leq n}\mu_{\mathcal{Q}}(z_{i}z_{ j})}{n}\), where \(\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{n}\). Moreover, since:  
\[\left(\sum_{i=1}^{n}\lambda_{i}\right)^{2}=\sum_{i=1}^{n}|\lambda_{i}|^{2}+2 \sum_{1\leq i<j\leq n}\lambda_{i}\lambda_{j} \tag{2}\]6  
\[\left(\sum_{i=1}^{n}\lambda_{i}\right)^{2}=\sum_{i=1}^{n}|\lambda_{i}|^{2}+2 \sum_{1\leq i<j\leq n}\lambda_{i}\lambda_{j} \tag{2}\]7  
Applying the Cauchy-Schwarz inequality to the vectors \((1,1,\ldots,1)\) and \((|\lambda_{1}|,|\lambda_{2}|,\ldots,|\lambda_{n}|)\) with \(n-1\) entries, we get:  
\[\left(\sum_{i=1}^{n}\lambda_{i}\right)^{2}=\sum_{i=1}^{n}|\lambda_{i}|^{2}+2 \sum_{1\leq i<j\leq n}\lambda_{i}\lambda_{j} \tag{2}\]8  
Substituting (5) in (6), we must have:  
\[\left(\sum_{i=1}^{n}\lambda_{i}\right)^{2}=\sum_{i=1}^{n}|\lambda_{i}|^{2}+2 \sum_{1\leq i<j\leq n}\lambda_{i}\lambda_{j} \tag{2}\]9  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]0  
Now, since the function:  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]1decreases on the interval:  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]2  
also \(n\leq 2\sum\limits_{1\leq i<j\leq n}(\mu_{Q}(z_{i}z_{j}))^{2}\), \(1\leq\frac{2\sum\limits_{1\leq i<j\leq n}(\mu_{Q}(z_{i}z_{j}))^{2}}{n}\). Therefore,  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]3  
Therefore, Equation (7) implies:  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]4  
Analogously, we can show that:  
\[\prod\limits_{i=1}^{n}(\lambda-\lambda_{i})=|A(\mathcal{G})-\lambda I|,\]5  
**Theorem 4**. : _Let \(\mathcal{G}=(\mathcal{P},\mathcal{Q})\) be a PFG on \(n\) vertices. Then, \(E(\mathcal{G})\leq\frac{n}{2}(1+\sqrt{n})\)._  
Proof. : Suppose that \(\mathcal{G}=(\mathcal{P},\mathcal{Q})\) is a PFG with \(n\) vertices. If \(n\leq 2\sum\limits_{1\leq i<j\leq n}(\mu_{Q}(z_{i}z_{j}))^{2}=2y\), then by routine calculus, it is easy to show that \(f(y)=\frac{2y}{n}+\sqrt{(n-1)(2y-(\frac{2y}{n}n)^{2})}\) is maximized when \(y=\frac{n^{2}+n\sqrt{n}}{4}\). Substituting this value of \(y\) in place of \(y=\sum\limits_{1\leq i<j\leq n}(\mu_{Q}(z_{i}z_{j}))^{2}\) in Theorem 3, we must have \(E(\mu_{Q}(z_{i}z_{j}))\leq\frac{n}{2}(1+\sqrt{n})\). Similarly, it is easy to show that \(E(\
u_{Q}(z_{i}z_{j}))\leq\frac{n}{2}(1+\sqrt{n})\). Hence, \(E(\mathcal{G})\leq\frac{n}{2}(1+\sqrt{n})\).