TITLE: Proving the inequlality $\int_{t_0}^{t_1}p(t)\exp(t_0-t)\leq \int_{t_0}^{t_1}p(t)\exp(t-t_1)$
QUESTION [2 upvotes]: Let $p$ be an increasing, continuous function on the interval $[t_0.t_1]$. I wish to show the inequality $$
\int_{t_0}^{t_1}p(t)\exp (t_0-t)dt\leq \int_{t_0}^{t_1}p(t)\exp(t-t_1)
$$
I have checked this inequality in the simple case that $p(t)=\sqrt{t},p(t)=t$ and $p(t)=t^2$, and it should be a simple inequality to proof but somehow it is giving me more struggles than it should.

REPLY [1 votes]: For the amended question, we can show the following generalisation:

Let $p$ and $q$ be increasing on $[a,b].$ Then $$ \int_a^b p(t) q(a+b-t) \mathrm{d}t\le \int_a^b p(t) q(t) \mathrm{d}t.$$

Proof. By rearranging the inequality, it suffices to argue that $$ \int_a^b (q(t) - q(a+b - t)) p(t) \mathrm{d}t \overset?\ge 0.$$
Now, to add some symmetry to our expressions, define $\mu = (a+b)/2, \delta = (b-a)/2,$ and make the substitution $t = z + \mu.$ We end up with the goal $$ \int_{-\delta}^\delta (q(\mu + z) - q(\mu - z)) p(z + \mu) \mathrm{d}z \overset?\ge 0,$$ where $p, q$ are increasing with $z$ in its range.
But observe by substituting $z \mapsto -z$ that $$ \int_{-\delta}^0 (q(\mu + z) - q(\mu - z)) p(z + \mu)  \mathrm{d}z
 = \int_0^\delta - (q(\mu + z) - q(\mu - z)) p(-z + \mu) \mathrm{d}z,$$ and so it suffices to argue that $$ \int_0^\delta (p (\mu + z) - p(\mu - z)) (q ( \mu + z) - q(\mu-z)) \mathrm{d}z\overset?\ge 0.$$ But this is obvious: since both $p$ and $q$ are increasing, and since $z \ge 0,$ the integrand above is positive. QED

Thanks to @William M. for helpful comments. They also point out that the statement holds if $p$ and $q$ are both decreasing as well, which follows since if $p$ and $q$ were decreasing, then $-p$ and $-q$ would be increasing, and $p q = -p \cdot -q$
Further, if one of $p$ and $q$ is increasing and the other decreasing, then applying the statement to one of $(p,-q)$ or $(-p,q)$ yields that $\int_a^b p(t) q(a+b - t) \ge \int_a^b p(t) q(t).$
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