Let $X$ be a positive, continuous random variable. Denote the density of $X$ by $f(x)$ and its CDF by $F(x)$. Let $\bar{F}(x) = 1- F(x)$ be the survival function of $X$. Given that the Hazard Rate, \begin{align} \lambda(x) &= \frac{f(x)}{\bar{F}(x)} \end{align} is increasing, i.e. $\lambda'(x) \geq 0$, I want to prove that \begin{align} \mathbf{E}\left[ X-c \mid X>c \right] &\leq \mathbf{E}\left[ X \right] \end{align} for any constant $c >0$.
Here's what I have tried so far:
Let \begin{align} \Lambda(x) &= \int_0^x \lambda(s) \mathop{}\!\mathrm{d} s. \end{align} This implies the following \begin{align} \lambda(x) &= \Lambda'(x), \\ -\Lambda(x) &= \log \left(\bar{F}(x) \right), \\ \bar{F}(x) &= e^{-\Lambda(x) }. \end{align} Writing the expected values as integrals (using the Darth Vader rule) and using the above, I can rewrite the inequality as \begin{align} \int_c^\infty e^{-\int_c^s \lambda(x) \mathop{}\!\mathrm{d}x } \mathop{}\!\mathrm{d}s &\leq \int_0^\infty e^{ -\int_0^s \lambda(x) \mathop{}\!\mathrm{d}x } \mathop{}\!\mathrm{d}s. \end{align} However, I do not see how this holds given $\lambda$ in increasing, and I can't think of a way using that assumption.
Since $\lambda$ is increasing, we have $\lambda(y+c) \geq \lambda(y)$, hence $\int_0^t \lambda(y + c) \,dy \geq \int_0^t \lambda(y)\,dy$, and thus $e^{-\int_0^t \lambda(y + c) \,dy} \leq e^{-\int_0^t \lambda(y) \,dy}$. Changing variables and using this inequality gives $$\int_c^\infty e^{-\int_c^s \lambda(x) \,dx} \,ds = \int_0^\infty e^{-\int_c^{t+c} \lambda(x) \,dx} \,dt = \int_0^\infty e^{-\int_0^t \lambda(y+c) \,dy} \,dt \leq \int_0^\infty e^{-\int_0^t \lambda(y) \,dy} \,dt $$ as desired.