Consider, $t:[0,\infty) \to \mathbb{R}$ given by $$ t(\lambda) = \text{argmin}_x\{f(x)+\lambda g(x)\}, \quad \quad\lambda \geq 0 $$ which is assumed to be welldefined, where $f$ and $g$ are positive and strictly convex and convex respectively (not that I think it matters for my question).
My intuition and also computer verifications, tells me that $f(t(\lambda))$ and $g(t(\lambda))$ are weakly increasing and weakly decreasing respectively.
How should I proceed to actually show this?
Take $\lambda_1< \lambda_2$, set $t_i:=t(\lambda_i)$. Then by optimality $$ f (t_1) +\lambda_1 g(t_1) \le f (t_2) +\lambda_1 g(t_2) = f (t_2) +\lambda_2 g(t_2) + (\lambda_1 - \lambda_2) g(t_2) \\ \le f(t_1) + \lambda_2 g(t_1) + (\lambda_1 - \lambda_2) g(t_2). $$ Hence $g$ is decreasing: $$ (\lambda_1 - \lambda_2) (g(t_2)-g(t_1))\ge0. $$ If $\lambda_1=0$, we have also $f(t_1) \le f(t_2)$.
Assume now $\lambda_1>0$. Then we can repeat the above arguments for minimization of $\lambda^{-1} f + g$. And we find the inequality $$ (\lambda_1^{-1} - \lambda_2^{-1})(f(t_2)-f(t_1))\ge0, $$ which means that $t\mapsto f(\lambda(t))$ is increasing.