The Grover's operator $G$ in Quantum Computing, has the following effect $$ G|\Phi\rangle=(4\sin^2\Delta -1)|\Phi\rangle-2\sin\Delta|z\rangle $$
where $|\Phi\rangle=\frac{1}{\sqrt{2^n}}(|x_1\rangle+|x_2\rangle+\cdots+|x_i\rangle+|z\rangle+|x_j\rangle+\cdots+|x_N\rangle)$ and $|x_k\rangle$ and $|z\rangle$ are orthonormal basis states, $\Delta\neq 0$ is the angle between $|\Phi\rangle$ and the vector orthogonal to $|z\rangle$.
The problem is to find the probability of measuring $|z\rangle$ after the Grover iteration $G$, ie., $P_G=|\langle z|G|\Phi\rangle|^2$ and show that it is greater than $P=|\langle z|\Phi\rangle|^2=\dfrac{1}{2^n}$.
We need to prove $|\langle z|G|\Phi\rangle|^2>|\langle z|\Phi\rangle|^2=\dfrac{1}{2^n}$ $$ \langle z|G|\Phi\rangle=(4\sin^2\Delta -1)\langle z|\Phi\rangle-2\sin\Delta\langle z|z\rangle=(4\sin^2\Delta -1)\frac{1}{\sqrt{2^n}}-2\sin\Delta\\ \\ |\langle z|G|\Phi\rangle|^2=\frac{1}{2^n}\bigg[16\sin^4\Delta+1-8\sin^2\Delta\bigg]+4\sin^2\Delta-4\sin\Delta\Big(\frac{4\sin^2\Delta}{\sqrt{2^n}}-1\Big)\\ \\ =\frac{1}{2^n}+\frac{16\sin^4\Delta}{2^n}+\sin^2\Delta\Big(4-\frac{8}{2^n}\Big)+\sin\Delta\Big(4-\frac{16}{\sqrt{2^n}}\sin^2\Delta\Big) $$
We have $\dfrac{16\sin^4\Delta}{2^n}> 0$
If I take $n$ to be very large then $4-\dfrac{8}{2^n}>0$ and therefore the term $\sin^2\Delta\Big(4-\frac{8}{2^n}\Big)> 0$
How do I show that the sum of the last 3 terms is positive ?
Reference : Exercise 52, Page 45, Grover’s Search Algorithm