# Why can the C5 formula also means that $\sum\limits_{k=1}^{K}tr(\mathbf F_{K4E})>P_T$?

In this paper:https://arxiv.org/pdf/1805.08898.pdf

I can understand why C4 means $$P_E > P^*$$,I just need to times $$1-\rho_k$$ in both two side,the C4 will become $$(1-\rho_k)[\sum\limits_{k=1}^{K}\mathbf h_k^H \mathbf F_j \mathbf h_k+\sigma^2_{\alpha_k}]\ge P$$,and it is the same as $$P_E > P^*$$,but why can C5 also mean $$\sum\limits_{k=1}^{K}tr(\mathbf F_{K4E})>P_T$$?In here,you can see $$F_{K4E}$$ as $$F_K$$.

$$P_T$$ is the power budget,it means that i have only $$P_T$$ power to send the message out

$$\sum\limits_{k=1}^{K}tr(\mathbf F_{K4E})$$ means that the total power i need to send the power

C5 comes from this formula: