Let $g(x)$ be a convex function and let $m \in R$. Show that the function $h(x)$ defined by $h(x) = \max(g(x),m)$ is convex.
The definition of convexity is that $f(x)$ is convex for $x_1, x_2 \in R$ and $\lambda_1, \lambda_2 > 0$ s.t. $\lambda_1 + \lambda_2 = 1$ iff $$f(\lambda_1x_1 + \lambda_2x_2) \le \lambda_1f(x_1) + \lambda_2f(x_2)$$
So I know that if $h(x)$ is to be convex then $$h(\lambda_1x_1 + \lambda_2x_2) \le \lambda_1h(x_1) + \lambda_2h(x_2)$$ $$\max(g(\lambda_1x_1 + \lambda_2x_2),m) \le \lambda_1\max(g(x_1),m) + \lambda_2\max(g(x_2),m)$$
But this is where I am stuck. Should I break the proof into different cases based on different values of $m$? I feel like I'm very close to this proof but can't see the key step.
Thanks in advance for your help.