Clearly $\sqrt{2f(x)} \ge 2\implies f(x)\ge 2$. Now, we multiply both sides by the conjugate to obtain \[ f(2x) \ge 2\sqrt{2f(x)} + 2\sqrt{2f(x)-f(2x)}\ge 2\sqrt{2f(x)} \]for all $x$. Now that $f(x)\ge 2$ for all $x$, we get that $f(2x)\ge 2\sqrt{2\cdot 2} = 4$. As $x$ is arbitrary, this concludes (a). Now, (b) is more interesting; I claim any $z<8$ works. To that end, note that $f(x)\ge 4$ yields $f(2x)\ge 2\sqrt{2\cdot 4} = 4\sqrt{2}$. This motivates to construct the sequence $(k_n)_{n\ge 1}$ defined by $k_1=2$ and for $n\ge 1$, \[ k_{n+1} = \frac{k_n+3}{2}. \]Note that $f(x)\ge 2^{k_t}$ for all $t$. Indeed, if $f(x)\ge 2^{k_n}$ then \[ f(2x) \ge 2\sqrt{2^{k_n+1}} = 2^{\frac{k_n+3}{2}} = 2^{k_{n+1}} \implies f(x)\ge 2^{k_{n+1}} \]as $x$ is arbitrary. Now I claim $k_n= 3-2^{1-n}$. Indeed, if $v_n =k_n-3$ the $2v_{n+1}=v_n$, yielding $v_n=2^{-n+1}v_1$ via a telescoping argument and thus $k_n = 3-2^{-n+1}$. With this, we get \[ f(x) \ge 2^{3-\frac{2}{2^n}} \]for all $n$. It is clear that for $n$ sufficiently large, this quantity dominates 7, yielding (b).

Can the bound be improved or equality can hold for at least one/infinitely many $x$?