# A Relaxed Inertial Method for Solving Monotone Inclusion Problems with Applications

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## Abstract

**:**

## 1. Introduction

## 2. Preliminaries

- The set of zeros of A is zer $A:=\{x\in \mathcal{H}\mid 0\in Ax\}$.
- The domain of A is dom $A:=\{x\in \mathcal{H}\mid Ax\ne \u2300\}$.
- The range of A is ran $A:=\{y\in \mathcal{H}\mid \exists \phantom{\rule{0.166667em}{0ex}}x\in \mathcal{H}:y\in Ax\}$.
- The graph set A is gra $A:=\{(x,y)\in {\mathcal{H}}^{2}\mid y\in Ax\}$.

- A is characterized as monotone if it satisfies the inequality $\langle x-y,u-v\rangle \ge 0$ for all $(x,u)$ and $(y,v)$ belonging to the graph of A.
- A is called maximal monotone if no other monotone operator $B:\mathcal{H}\to {2}^{\mathcal{H}}$ exists for which its graph strictly encompasses the graph of A.
- A is called $\beta $-strongly monotone with $\beta \in (0,+\infty )$ if for all $(x,u)\in \mathrm{gra}\phantom{\rule{0.166667em}{0ex}}A$ and $(y,v)\in \mathrm{gra}\phantom{\rule{0.166667em}{0ex}}A$, there holds that $\langle x-y,u-v\rangle \ge {\beta \parallel x-y\parallel}^{2}$.
- A is said to be $\beta $-cocoercive with $\beta \in (0,+\infty )$ if ${\beta \parallel Ax-Ay\parallel}^{2}\le \langle Ax-Ay,x-y\rangle $ for all $x,y\in \mathcal{H}$.
- The resolvent of A is defined by$${J}_{\lambda A}={(Id+\lambda A)}^{-1},$$

**Lemma**

**1**

- (i)
- ${\sum}_{k=1}^{\infty}{[{\phi}_{k}-{\phi}_{k-1}]}_{+}<+\infty $, where ${\left[t\right]}_{+}=max\{t,0\}$;
- (ii)
- There exists ${\phi}^{*}\in [0,+\infty )$ such that ${lim}_{k\to +\infty}{\phi}_{k}={\phi}^{*}$.

**Lemma**

**2**

- (1)
- For all ${x}^{*}\in \mathcal{C}$, ${lim}_{k\to \infty}\parallel {x}_{k}-{x}^{*}\parallel $ exists;
- (2)
- Every weak sequential cluster point of ${\left\{{x}_{k}\right\}}_{k\ge 1}$ belongs to $\mathcal{C}$.

## 3. The RIFBHF Algorithm

**Remark**

**1.**

- (i)
- FBHF method [15]: assume ${\alpha}_{k}=0$ and ${\lambda}_{k}=1$ when $k\ge 1$,$$\left\{\begin{array}{c}{x}_{k}={J}_{{\gamma}_{k}A}({z}_{k}-{\gamma}_{k}(B+C){z}_{k}),\hfill \\ {z}_{k+1}={x}_{k}-{\gamma}_{k}(B{x}_{k}-B{z}_{k}).\hfill \end{array}\right.$$
- (ii)
- Inertial forward–backward–half-forward scheme [40]: assume ${\lambda}_{k}=1$ when $k\ge 1$,$$\left\{\begin{array}{c}{w}_{k}={z}_{k}+{\alpha}_{k}({z}_{k}-{z}_{k-1}),\hfill \\ {x}_{k}={J}_{{\gamma}_{k}A}({w}_{k}-{\gamma}_{k}(B+C){w}_{k}),\hfill \\ {z}_{k+1}={x}_{k}+{\gamma}_{k}(B{w}_{k}-B{x}_{k}).\hfill \end{array}\right.$$
- (iii)
- Relaxed forward–backward–half-forward method: assume ${\alpha}_{k}=0$ when $k\ge 1$,$$\left\{\begin{array}{c}{x}_{k}={J}_{{\gamma}_{k}A}({z}_{k}-{\gamma}_{k}(B+C){z}_{k}),\hfill \\ {t}_{k}={x}_{k}-{\gamma}_{k}(B{x}_{k}-B{z}_{k}),\hfill \\ {z}_{k+1}=(1-{\lambda}_{k}){z}_{k}+{\lambda}_{k}{t}_{k}.\hfill \end{array}\right.$$

**Proposition**

**1.**

**Proof.**

**Proposition**

**2.**

- (i)
- $\parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}\le \parallel {w}_{k}-{z}^{*}{\parallel}^{2}-{\phi}_{k}{\parallel {z}_{k+1}-{w}_{k}\parallel}^{2},$where ${\phi}_{k}=\left(\frac{{L}^{2}({\chi}^{2}-{\gamma}_{k}^{2})}{{\lambda}_{k}{(1+{\gamma}_{k}L)}^{2}}+\frac{1-{\lambda}_{k}}{{\lambda}_{k}}\right)$.
- (ii)
- Define that$${\Pi}_{k}:=\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}+\left({\alpha}_{k}(1+{\alpha}_{k})-{\phi}_{k}({\alpha}_{k}^{2}-{\alpha}_{k})\right){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}.$$

**Proof.**

- (i)
- Proposition 1 leads to$$\begin{array}{c}\parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}\hfill \\ =\parallel (1-{\lambda}_{k}){w}_{k}+{\lambda}_{k}{t}_{k}-{z}^{*}{\parallel}^{2}\hfill \\ =\parallel (1-{\lambda}_{k})({w}_{k}-{z}^{*})+{\lambda}_{k}({t}_{k}-{z}^{*}){\parallel}^{2}\hfill \\ =(1-{\lambda}_{k})\parallel {w}_{k}-{z}^{*}{\parallel}^{2}+{\lambda}_{k}\parallel {t}_{k}-{z}^{*}{\parallel}^{2}-{\lambda}_{k}(1-{\lambda}_{k}){\parallel {t}_{k}-{w}_{k}\parallel}^{2}\hfill \\ \le (1-{\lambda}_{k}){\parallel {w}_{k}-{z}^{*}\parallel}^{2}+{\lambda}_{k}\left(\parallel {w}_{k}-{z}^{*}{\parallel}^{2}-{L}^{2}({\chi}^{2}-{\gamma}_{k}^{2}){\parallel {w}_{k}-{x}_{k}\parallel}^{2}\right)\hfill \\ \phantom{\rule{2.em}{0ex}}\phantom{\rule{1.em}{0ex}}-{\lambda}_{k}(1-{\lambda}_{k}){\parallel {t}_{k}-{w}_{k}\parallel}^{2}\hfill \\ =\parallel {w}_{k}-{z}^{*}{\parallel}^{2}-{\lambda}_{k}{L}^{2}({\chi}^{2}-{\gamma}_{k}^{2})\parallel {w}_{k}-{x}_{k}{\parallel}^{2}-{\lambda}_{k}(1-{\lambda}_{k}){\parallel {t}_{k}-{w}_{k}\parallel}^{2}.\hfill \end{array}$$According to the Lipschitz continuity of B,$$\begin{array}{cc}\hfill \frac{1}{{\lambda}_{k}}\parallel {z}_{k+1}-{w}_{k}\parallel & =\parallel {t}_{k}-{w}_{k}\parallel \le \parallel {t}_{k}-{x}_{k}\parallel +\parallel {x}_{k}-{w}_{k}\parallel \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& ={\gamma}_{k}\parallel B{w}_{k}-B{x}_{k}\parallel +\parallel {w}_{k}-{x}_{k}\parallel \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \le (1+{\gamma}_{k}L)\parallel {w}_{k}-{x}_{k}\parallel ,\hfill \end{array}$$$$\frac{{L}^{2}\left({\chi}^{2}-{\gamma}_{k}^{2}\right)}{{\lambda}_{k}{(1+{\gamma}_{k}L)}^{2}}\parallel {z}_{k+1}-{w}_{k}{\parallel}^{2}\le {\lambda}_{k}{L}^{2}\left({\chi}^{2}-{\gamma}_{k}^{2}\right){\parallel {w}_{k}-{x}_{k}\parallel}^{2}.$$Combining (13) and (14), we have$$\parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}\le \parallel {w}_{k}-{z}^{*}{\parallel}^{2}-\left(\frac{{L}^{2}\left({\chi}^{2}-{\gamma}_{k}^{2}\right)}{{\lambda}_{k}{(1+{\gamma}_{k}L)}^{2}}+\frac{1-{\lambda}_{k}}{{\lambda}_{k}}\right){\parallel {z}_{k+1}-{w}_{k}\parallel}^{2}.$$
- (ii)
- It follows from the definition of ${w}_{k}$ and the Cauchy–Schwartz inequality that$$\begin{array}{cc}\hfill \parallel {z}_{k+1}-{w}_{k}{\parallel}^{2}& =\parallel {z}_{k+1}-{z}_{k}-{\alpha}_{k}({z}_{k}-{z}_{k-1}){\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}+{\alpha}_{k}^{2}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}-2{\alpha}_{k}\langle {z}_{k+1}-{z}_{k},{z}_{k}-{z}_{k-1}\rangle \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \ge \parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}+{\alpha}_{k}^{2}\parallel {z}_{k}-{z}_{k-1}{\parallel}^{2}-2{\alpha}_{k}\parallel {z}_{k+1}-{z}_{k}\parallel \parallel {z}_{k}-{z}_{k-1}\parallel \hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \ge (1-{\alpha}_{k})\parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}+({\alpha}_{k}^{2}-{\alpha}_{k}){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}.\hfill \end{array}$$Simultaneously, we have$$\begin{array}{cc}\hfill \parallel {w}_{k}-{z}^{*}{\parallel}^{2}& =\parallel {z}_{k}+{\alpha}_{k}({z}_{k}-{z}_{k-1})-{z}^{*}{\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =\parallel (1+{\alpha}_{k})({z}_{k}-{z}^{*})-{\alpha}_{k}({z}_{k-1}-{z}^{*}){\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =(1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}+{\alpha}_{k}(1+{\alpha}_{k}){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}.\hfill \end{array}$$By Propositions 2(i), (15), and (16), we have$$\begin{array}{cc}\hfill \parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}& \le \parallel {w}_{k}-{z}^{*}{\parallel}^{2}-{\phi}_{k}{\parallel {z}_{k+1}-{w}_{k}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \le (1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}+{\alpha}_{k}(1+{\alpha}_{k}){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{1.em}{0ex}}\phantom{\rule{2.em}{0ex}}-{\phi}_{k}\left((1-{\alpha}_{k})\parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}+({\alpha}_{k}^{2}-{\alpha}_{k}){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\right)\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =(1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}-{\phi}_{k}(1-{\alpha}_{k}){\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{1.em}{0ex}}\phantom{\rule{2.em}{0ex}}+\left({\alpha}_{k}(1+{\alpha}_{k})-{\phi}_{k}({\alpha}_{k}^{2}-{\alpha}_{k})\right){\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =(1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}-{\pi}_{k}{\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{1.em}{0ex}}\phantom{\rule{2.em}{0ex}}+{\eta}_{k}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2},\hfill \end{array}$$$${\Pi}_{k}:=\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}+{\eta}_{k}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}.$$Now, by (17) and ${\alpha}_{k}\le {\alpha}_{k+1}$, we obtain$$\begin{array}{cc}\hfill {\Pi}_{k+1}-{\Pi}_{k}& =\parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}-{\alpha}_{k+1}\parallel {z}_{k}-{z}^{*}{\parallel}^{2}+{\eta}_{k+1}{\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{2.em}{0ex}}-\parallel {z}_{k}-{z}^{*}{\parallel}^{2}+{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}-{\eta}_{k}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \le \parallel {z}_{k+1}-{z}^{*}{\parallel}^{2}-(1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}+{\alpha}_{k}{\parallel {z}_{k-1}-{z}^{*}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{2.em}{0ex}}+{\eta}_{k+1}\parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}-{\eta}_{k}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \le (1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}-{\alpha}_{k}\parallel {z}_{k-1}-{z}^{*}{\parallel}^{2}-{\pi}_{k}{\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{2.em}{0ex}}+{\eta}_{k}\parallel {z}_{k}-{z}_{k-1}{\parallel}^{2}-(1+{\alpha}_{k})\parallel {z}_{k}-{z}^{*}{\parallel}^{2}+{\alpha}_{k}{\parallel {z}_{k-1}-{z}^{*}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \phantom{\rule{2.em}{0ex}}+{\eta}_{k+1}\parallel {z}_{k+1}-{z}_{k}{\parallel}^{2}-{\eta}_{k}{\parallel {z}_{k}-{z}_{k-1}\parallel}^{2}\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& =-({\pi}_{k}-{\eta}_{k+1}){\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}.\hfill \end{array}$$If follows from $0<{\alpha}_{k}\le {\alpha}_{k+1}\le \alpha $ that$$\begin{array}{cc}\hfill {\pi}_{k}-{\eta}_{k+1}& ={\phi}_{k}(1-{\alpha}_{k})-{\alpha}_{k+1}(1+{\alpha}_{k+1})+{\phi}_{k+1}({\alpha}_{k+1}^{2}-{\alpha}_{k+1})\hfill \\ \hfill \phantom{\rule{1.em}{0ex}}& \ge {\phi}_{k}(1-{\alpha}_{k+1})-{\alpha}_{k+1}(1+{\alpha}_{k+1})+{\phi}_{k+1}({\alpha}_{k+1}^{2}-{\alpha}_{k+1}).\hfill \end{array}$$Let ${\zeta}_{k}={\phi}_{k}(1-{\alpha}_{k+1})-{\alpha}_{k+1}(1+{\alpha}_{k+1})+{\phi}_{k+1}({\alpha}_{k+1}^{2}-{\alpha}_{k+1})$; we obtain$${\Pi}_{k+1}-{\Pi}_{k}\le -{\zeta}_{k}{\parallel {z}_{k+1}-{z}_{k}\parallel}^{2}.$$The proof is completed.

**Theorem**

**1.**

**Proof.**

**Remark**

**2**

**Remark**

**3.**

**Remark**

**4.**

## 4. Composite Monotone Inclusion Problem

**Corollary**

**1.**

**Proof.**

**Corollary**

**2.**

**Proof.**

## 5. Numerical Experiments

- FBHF [15]: $\gamma =\frac{4\beta}{1+\sqrt{1+16{\beta}^{2}{L}^{2}}}$ for four image blurring scenarios.
- PD: the first-order primal–dual splitting algorithm [42] with $\tau =\frac{1}{3}$, $\sigma =\frac{1}{3}$, and $\theta =1$ for four image blurring scenarios.
- RIFBHF: $\gamma =\frac{4\beta}{1+\sqrt{1+16{\beta}^{2}{L}^{2}}}$ using $\alpha =0.2$ and $\lambda =0.9$ for image blurring scenarios 1 and 2 and $\alpha =0.6$ and $\lambda =0.8$ for image blurring scenarios 3 and 4.

## 6. Conclusions

## Author Contributions

## Funding

## Data Availability Statement

## Acknowledgments

## Conflicts of Interest

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**Figure 1.**Balance between ${\alpha}_{k}$ and ${\lambda}_{k}$ with $\epsilon =0.1$, $\gamma =\frac{0.45}{L}$ (

**left**), and $\gamma =\frac{0.9}{L}$ (

**right**).

**Figure 2.**Behaviour of the error of objective value $log(f-{f}^{*})/{f}^{*}$ against running time for different parameters $\alpha $, $\gamma $, and $\lambda $.

**Figure 3.**Behaviour of SNR against running time for different parameters $\alpha $, $\gamma $, and $\lambda $.

**Figure 4.**Behaviour of the error of objective value $log(f-{f}^{*})/{f}^{*}$ against running time for different image burring scenarios, i.e., (

**a**) scenario 1, (

**b**) scenario 2, (

**c**) scenario 3, and (

**d**) scenario 4.

**Figure 5.**(

**a**) The original image of Barbara. (

**b**) The blurred image of Barbara. (

**c**) The deblurred image by FBHF. (

**d**) The deblurred image by RIFBHF.

**Table 1.**The parameter selection range for FBHF [15] and RIFBHF.

Algorithms | ${\mathit{\gamma}}_{\mathit{k}}$ | ${\mathit{\alpha}}_{\mathit{k}}$ | ${lim}_{\mathit{k}\to +\mathit{\infty}}{\mathit{\lambda}}_{\mathit{k}}=\mathit{\lambda}$ |
---|---|---|---|

FBHF [15] | $[\eta ,\chi -\eta ]$ | 0 | 1 |

RIFBHF | $[\eta ,\chi -\eta ]$ | $[0,1)$ | $\left(0,\frac{2(1+\gamma L)-\epsilon}{{(1+\gamma L)}^{2}}\frac{{(1-\alpha )}^{2}}{(2{\alpha}^{2}-\alpha +1)}\right)$ |

Scenario | Blur Kernel | Gaussian Noise |
---|---|---|

1 | $9\times 9$ box average kernel | $\sigma =1.5$ |

2 | $9\times 9$ box average kernel | $\sigma =3$ |

3 | $7\times 7$ Gaussian kernel with ${\sigma}_{a}=10$ | $\sigma =1.5$ |

4 | $7\times 7$ Gaussian kernel with ${\sigma}_{a}=10$ | $\sigma =3$ |

**Table 3.**SNR values and iterations when $\gamma =\frac{4}{1+\sqrt{1+16\ast 8}}$, $\alpha =0$, and $\lambda =1$ with different parameter $\mu $ for Barbara image in four blurred scenarios.

Scenario | 1 | 2 | 3 | 4 | |||||
---|---|---|---|---|---|---|---|---|---|

$\mathit{\mu}$ | SNR (dB) | Iter | SNR (dB) | Iter | SNR (dB) | Iter | SNR (dB) | Iter | |

$0.1$ | $17.5358$ | 50 | $17.5144$ | 52 | $17.9741$ | 45 | $17.9510$ | 48 | |

$0.5$ | $17.5414$ | 55 | $17.5196$ | 56 | $17.9385$ | 49 | $17.9148$ | 51 | |

1 | $17.5515$ | 61 | $17.5304$ | 62 | $17.9071$ | 54 | $17.8799$ | 55 | |

2 | $17.4859$ | 65 | $17.4658$ | 65 | $17.8012$ | 56 | $17.7788$ | 57 | |

3 | $17.4001$ | 65 | $17.3831$ | 65 | $17.7126$ | 56 | $17.6937$ | 57 | |

4 | $17.3251$ | 65 | $17.3106$ | 66 | $17.6368$ | 57 | $17.6203$ | 57 | |

5 | $17.2584$ | 67 | $17.2443$ | 68 | $17.5713$ | 58 | $17.5553$ | 59 | |

6 | $17.1948$ | 69 | $17.1823$ | 69 | $17.5117$ | 60 | $17.4972$ | 60 | |

7 | $17.1344$ | 71 | $17.1230$ | 71 | $17.4563$ | 61 | $17.4422$ | 62 | |

8 | $17.0779$ | 73 | $17.0663$ | 74 | $17.4033$ | 63 | $17.3895$ | 64 | |

9 | $17.0237$ | 75 | $17.0123$ | 76 | $17.3530$ | 65 | $17.3413$ | 65 | |

10 | $16.9718$ | 77 | $16.9608$ | 78 | $17.3049$ | 67 | $17.2940$ | 67 |

**Table 4.**SNR values and iterations when $\gamma =\frac{4}{1+\sqrt{1+16\ast 8}}$, $\alpha =0.3$, and $\lambda =0.6$ with different parameter $\mu $ for Barbara image in four blurred scenarios.

Scenario | 1 | 2 | 3 | 4 | |||||
---|---|---|---|---|---|---|---|---|---|

$\mathit{\mu}$ | SNR (dB) | Iter | SNR (dB) | Iter | SNR (dB) | Iter | SNR (dB) | Iter | |

$0.1$ | $17.4757$ | 52 | $17.4513$ | 53 | $17.9109$ | 46 | $17.8957$ | 49 | |

$0.5$ | $17.4740$ | 55 | $17.4573$ | 56 | $17.8890$ | 50 | $17.8643$ | 51 | |

1 | $17.5060$ | 62 | $17.4884$ | 63 | $17.8783$ | 56 | $17.8545$ | 57 | |

2 | $17.4687$ | 68 | $17.4502$ | 68 | $17.7913$ | 59 | $17.7707$ | 60 | |

3 | $17.3929$ | 69 | $17.3785$ | 70 | $17.7093$ | 60 | $17.6920$ | 61 | |

4 | $17.3214$ | 70 | $17.3072$ | 70 | $17.6367$ | 61 | $17.6212$ | 61 | |

5 | $17.2577$ | 72 | $17.2446$ | 72 | $17.5723$ | 62 | $17.5582$ | 63 | |

6 | $17.1970$ | 74 | $17.1850$ | 75 | $17.5144$ | 64 | $17.5007$ | 65 | |

7 | $17.1384$ | 76 | $17.1277$ | 76 | $17.4600$ | 66 | $17.4480$ | 66 | |

8 | $17.0832$ | 78 | $17.0725$ | 79 | $17.4078$ | 68 | $17.3953$ | 69 | |

9 | $17.0298$ | 81 | $17.0203$ | 81 | $17.3580$ | 70 | $17.3463$ | 71 | |

10 | $16.9787$ | 83 | $16.9686$ | 84 | $17.3103$ | 72 | $17.3005$ | 72 |

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**MDPI and ACS Style**

Zong, C.; Tang, Y.; Zhang, G.
A Relaxed Inertial Method for Solving Monotone Inclusion Problems with Applications. *Symmetry* **2024**, *16*, 466.
https://doi.org/10.3390/sym16040466

**AMA Style**

Zong C, Tang Y, Zhang G.
A Relaxed Inertial Method for Solving Monotone Inclusion Problems with Applications. *Symmetry*. 2024; 16(4):466.
https://doi.org/10.3390/sym16040466

**Chicago/Turabian Style**

Zong, Chunxiang, Yuchao Tang, and Guofeng Zhang.
2024. "A Relaxed Inertial Method for Solving Monotone Inclusion Problems with Applications" *Symmetry* 16, no. 4: 466.
https://doi.org/10.3390/sym16040466