Suppose $f$ is a continuous decreasing (or more generally, nonincreasing) function on $(a,b)$ and $F$ is an antiderivative of $f$ there. For $0 < t < 1$ and $a < x < y < b$, we need to prove that $F(tx + (1-t) y) \ge t F(x) + (1-t) F(y)$. Rearrange this as $ t \left(F(m) - F(x)\right) \ge (1-t) \left( F(y) - F(m) \right)$, where $m = tx + (1-t) y$. Now by the Mean Value Theorem, $F(m) - F(x) = (m-x) f(u)$ for some $u$ between $x$ and $m$, while $F(y) - F(m) = (y-m) f(v)$ for some $v$ between $m$ and $y$. Note that $u \le v$ so $f(u) \ge f(v)$, and $m - x = (1-t)(y-x)$ while $y-m = t (y-x)$. So indeed $t \left(F(m) - F(x) \right) = t(1-t) (y-x) f(u) \ge t(1-t)(y-x) f(v) = (1-t) \left(F(y) - F(m)\right)$