If $L$ is a positive semi-definite matrix and all elements in each row add up to 0, How to prove that the inf-norm of $(I+L)(I+L+L^2)^{-1} \leq 1$?

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Assuming that the L is a positive semi-definite matrix and $$ \sum\limits_{j=1}^{N} L_{i,j} = 0. \quad for \; i = 1,...,N.$$ I want to prove that $\|(I+L) (I+L+L^2)^{-1}\|_{\infty} \leq 1$, I have founded the proof of $\|(I+L)^{-1}\|_{\infty} \leq 1$ requires an additional diagonal dominance condition.

I have try to prove the $\|(I+L)^{-1}\|_{\infty} \leq 1$ by Diagonalization,i.e, there exists an invertible matrix, denoted as P, such that: $L=P^{-1} \Lambda P, \Lambda = diag(\lambda_1,...,\lambda_N)$ and $\lambda_i \geq 0$. So I can rewrite $(I+L)^{-1} = (I + P^{-1} \Lambda P )^{-1}=(P^{-1}(I + \Lambda) P)^{-1} = P^{-1} (I+\Lambda)^{-1} P$, consequently, $\|(I+L)^{-1}\|_{\infty} \leq \|P^{-1}\|_{\infty} \|(I+\Lambda)^{-1}\|_{\infty}\|P\|_{\infty} \leq\|P^{-1}\|_{\infty} \|P\|_{\infty} \leq 1$. This proof process does not seem to require that L is diagonally dominant, am I missing any steps? Whether it is eventually less than $1$ will be determined by the condition number of P, and it can be easily extended to proof $\left\|(I+L)(I+L+L^2)^{-1}\right\|_{\infty} \leq 1$?

Is there a theory about the matrix P composed of eigenvectors that can demonstrate it? If the infinity norm cannot be proven, can the 2-norm be proven? I would be very grateful if you could provide a reference to the proof of this claim.

EDIT 1: The last step $\|P^{-1}\| \|P\| \leq 1$ does not hold for inf-norm but holds for 2-norm. Is there any ways could proof it for the case of inf-norm?

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I assume that $\|\cdot\|_{\infty}$ means the spectral norm, then it is equivalent to prove that $\sigma_{min}$ of the inverse is no less than $1$, but this is somehow clear since the inverse is simply $$(I+L+L^2)(I+L)^{-1}=I+L^2(I+L)^{-1},$$ and by diagonalizing $L$ or writing it into norm form we can see that $L^2(I+L)^{-1}$ is still positive semi-definte.