This seem hard.....

Let $P(x,y,z)$ denote the assertion in the problem. $P(0,0,0) \Rightarrow f(f(0))=0$ Let us set $k=f(0)$. $P(k,k,y) \Rightarrow 2y \ge f(0)+2 \max \{ k,f(y) \} -k \ge f(0)+k$ By setting $y$ small enough one can see $f(0)=0$.Let $Q(y,z)$ denote the assertion $P(0,y,z)$. $Q(0,x) \Rightarrow f(x) \ge x \forall x \in \mathbb{R}$ $Q(x,0) \Rightarrow f(x) \le x \forall x \in \mathbb{R}$ So $f(x)=x \forall x \in \mathbb{R}$

Let $P(x,y,z)$ denote the assertion in the problem. $P(0,0,0) \Rightarrow 2f(0) \ge 2f(0) + f(f(0)) \Rightarrow f(f(0)) = 0$ $P(f(0),0,f(0)) \Rightarrow 0 \ge f(0) \Rightarrow f(0)=0$ $P(x,x,x) \Rightarrow x \ge f(f(x)) *$ $P(0,f(0),x) \Rightarrow f(x) \ge x **$ $* , ** \Rightarrow f(x) = x$

$P(x,y,z)\implies x+2 \max\{y,f(x),f(z)\} \ge f(f(x))+2 \max\{z,f(y)\}$ $P(0,0,0)\implies f(f(0))=0$ $P(f(0),0,f(0))\implies f(0)=0$ $P(0,0,x)\implies f(x)\geq x$ $P(0,x,0)\implies x\geq f(x)$ $\therefore f(x)\equiv x \text{ (works)}$

Let $P(x, y, z)$ denote the assertion. $P(0,0,0): 2\max(0,f(0))\geqslant f(f(0))+2\max(0,f(0)) \Rightarrow 0 \geqslant f(f(0)).$ But $f$ ranges over the nonnegatives, so $f(f(0))=0$. $P(f(0), 0, f(0)): f(0) \geqslant 3f(0) \Rightarrow f(0)=0$ $P(0, 0, x>0): 2\max(0, f(x)) \geqslant 2x \Rightarrow \max(0, f(x)) \geqslant x$ $P(0, x>0, 0): 2\max(x, 0) \geqslant 2\max(0, f(x)) \Rightarrow x \geqslant \max(0, f(x))$ Combine the last two, and $\max(0, f(x))=x$. If $f(x)=0$ then $x=0$, but $x>0$, contradiction. Once again $f$ ranges over the nonnegatives, so otherwise, $f(x)>0$ and therefore $f(x)=x.$

Let $P(x,y,z)$ denote the assertion. $P(x,f(x),x): x+2f(x) \ge f(f(x)) + 2\max\{x,f(f(x))\}\ge f(f(x))+2x \quad (2)$ Then $f(f(x))\le 2f(x)-x \quad (1)$ Fix $x>0$ and consider the sequence $u_0 = x, u_1=f(x), u_{n+1} = f(u_n) \quad \forall n\ge 1$ then $u_n \ge 0 \quad \forall n$ From (1) we have $$u_{n+1} \le 2u_n - u_{n-1}$$So we have $$u_{n+1} - u_n \le u_n - u_{n-1} \le \ldots u_1 - u_0 = f(x)-x$$then we have $$0 \le u_{n+1} \le u_n + f(x)-x \le u_{n-1} + 2(f(x)-x) \le u_1 + n(f(x)-x)$$So $$f(x)-x \ge \dfrac{-u_1}{n}$$Let $n\rightarrow +\infty$ we have $f(x) \ge x \quad \forall x \ge 0$ From $(2)$ we have $x+2f(x) \ge 3f(f(x))\ge 3f(x)$ then $x\le f(x) \le x$ so $f(x) = x \quad \forall x \ge 0$. We can check that this function satisfied the requirement.