Let $\phi:\mathbb [0,\infty) \to [0,\infty)$ be a $C^2$ function, and let $c>0$ be a constant.
Suppose that for any $x_1,x_2>0, \alpha \in [0,1]$ satisfying $\alpha x_1 + (1- \alpha)x_2 =c$, we have
$$
\phi(c)=\phi\left(\alpha x_1 + (1- \alpha)x_2 \right) \leq \alpha \phi(x_1) + (1-\alpha)\phi(x_2) \tag{1}
$$
Is it true that
$$
\phi(x) \ge \phi(c)+\phi'(c) (x-c) \tag{2}
$$
for every $x$ in a neighbourhood $c$?
Does the answer change if we assume in addition that $\phi$ is strictly decreasing?
The classical proofs for "convexity of $\phi$ implies $\phi$ is above its tangents do not seem to adapt to this setup.
The converse claim holds:
Indeed,
$$
\phi(x_1) \ge \phi(c)+\phi'(c) (x_1-c) , \phi(x_2) \ge \phi(c)+\phi'(c) (x_2-c)
$$
thus
$$
\alpha \phi(x_1) + (1-\alpha)\phi(x_2) \ge \phi(c)+\phi'(c)(\alpha x_1 + (1- \alpha)x_2 -c)=\phi(c).
$$
This formula here implies that $\phi''(c) \ge 0$.
Best Answer
It's true. The most general formulation would be like this: Suppose your condition (1) is satisfied, and $$\phi'_{-}(c)=\lim_{h\to 0^-}\frac{\phi(c+h)-\phi(c)}{h}$$ exists. Then, $$\phi(x) \ge \phi(c)+\phi'_{-}(c) (x-c) \tag{3}$$ for $x>c.$ If $$\phi'_{+}(c)=\lim_{h\to 0^+}\frac{\phi(c+h)-\phi(c)}{h}$$ exists, then $$\phi(x) \ge \phi(c)+\phi'_{+}(c) (x-c) \tag{4}$$ for $0<x<c.$
Proof: Let $x>c,$ first. Then, for $h>0,$ we can choose $\alpha$ so that $$c=\alpha\,(c-h)+(1-\alpha)\,x$$ i.e. $$\alpha=\frac{x-c}{x-c+h},\quad 1-\alpha=\frac{h}{x-c+h}.$$ With $x_1=c-h$ and $x_2=x,$ your condition (1) becomes (after some algebra) $$\phi(x)\ge(x-c)\,\frac{\phi(c)-\phi(c-h)}{h}+\phi(c),$$ and $h\to0$ proves (3). The proof of (4) (with $x_1=x<c$ and $x_2=c+h>c$) is completely analogous and left to the reader. ;-)