An elementary integral inequality I was playing around with some integrals and decided that
$$
a\int_a^b t^{n−2}\alpha(t) dt 
\le
\int_a^b t^{n−1}\alpha(t) dt
$$
for $a,b>0$, $\alpha \ge 0$, $\alpha$ nice enough so that the above expression is defined, should be true.
The idea is that the little $a$ on the left hand side compensates when $t<1$ (i.e. $a<1$).  One can produce a cases `brute force' style proof from this observation.
But, I wanted to ask the following: is the above inequality sharp?  Can the power of $t$ on the LHS be played with?  Can the constant $a$ be improved (perhaps depending on additional properties of $\alpha$)?
And most importantly: does this little guy have a name?
 A: Expanding and correcting my comment (although I am not sure to which extent this answers your question).
[Maybe this is closer to a comment as to an answer, but it is too long. More or less, this is still an attempt to get a more precise formulation of the question by exhibiting some special cases.]
First let us have a look at the situation when a single function $\alpha$ is considered.
If the function $\alpha$ has the property that there exists a subinterval $(x,y)\subseteq \langle a,b \rangle$ such that $\alpha$ is bounded from zero on this subinterval, i.e., there exists $\varepsilon>0$ such that $\alpha(t)>\varepsilon$ for each $t\in(x,y)$, then the above inequality is strict ($a$ is not the best possible constant).
$$\int_a^b t^{n−1}\alpha(t) dt-a\int_a^b t^{n−2}\alpha(t) dt=
\int_a^b (t-a)t^{n−2}\alpha(t) dt \ge $$
$$\int_x^y (t-a)t^{n−2}\alpha(t) dt \ge 
\varepsilon \int_x^y (t-a)t^{n−2} dt >0.$$
On the other hand, if we consider class of functions such that for each $\varepsilon>0$ there is some $\alpha$ with the property that $\alpha(t)\le 1$ for each $t$ and $\alpha(t)$ is zero for $t>a+\varepsilon$, then
$$\int_a^b (t-a)t^{n−2}\alpha(t) dt \le 
\int_a^{a+\varepsilon} t^{n-1}-at^{n-2} =$$
$$\left[\frac{t^n}n-a\frac{t^{n-1}}{n-1} \right]_a^{a+\varepsilon}=
\frac{(a+\varepsilon)^n-a^n}n - a\frac{(a+\varepsilon)^{n-1}-a^{n-1}}{n-1}.$$
Both fractions in the last expression tend to 0 as $\varepsilon\to 0^+$, so in this case, $a$ is the best possible constant. 
(I believe that similar reasoning would work not only for $\alpha(t)=0$ for $t>a+\varepsilon$, but also if it is sufficiently small for $t>a+\varepsilon$.)
A class of functions, which I did not address in this post and which might be interesting could be the class of continuous functions fulfilling $\int_a^b \alpha(t) dt=1$.
EDIT: Now I realized that the above approach works also for the functions fulfilling $\int_a^b \alpha(t) dt=1$.
Notice that for the step function
$$\alpha(t)=
  \begin{cases}
    \frac1\varepsilon & t\in [a,a+\varepsilon], \\
    0 & \text{otherwise}.
  \end{cases}
$$
we have $\int_a^b \alpha(t) dt=1$ and 
$$\int_a^b (t-a)t^{n-2}\alpha(t) dt =
\frac1\varepsilon \int_a^{a+\varepsilon} t^{n-1}-at^{n-2} =$$
$$\frac1\varepsilon \left[\frac{t^n}n-a\frac{t^{n-1}}{n-1} \right]_a^{a+\varepsilon}=
\frac 1n \frac{(a+\varepsilon)^n-a^n}\varepsilon - a \frac1{n-1} \frac{(a+\varepsilon)^{n-1}-a^{n-1}}{\varepsilon}.$$
If we notice that $\lim\limits_{\varepsilon\to0^+} \frac{(a+\varepsilon)^n-a^n}\varepsilon = na^{n-1}$ (derivative of the function $x^n$) then again the last expression converges to zero again.
Step functions can be approximated by continuous functions, so with some effort the function $\alpha$ can be made continuous. 
A: Another approach and somehow an answer to your second question: This can be shown by an application of the reverse Hölder inequality in the limiting case.
For $f,g$ with $g\neq 0$ almost everywhere we have for (if the right hand side exists)
$$\|f\|_1 = \||fg|\,|g|^{-1}|\|_1 \leq \|fg\|_1\||g|^{-1}\|_\infty = \|fg\|_1\inf|g|$$
(a more general case is treated here).
However, I do not know about sharpness or optimal constants here, although it seems that there is literature on this..
A: I imagine you would want the inequality to hold for all $b>a$.  Suppose $f(n)$ is a candidate to replace the $n-2$ in the LHS, and $A=A(a,b,\alpha,n)^{\ddagger}$ is a candidate to replace the $a$ out front:  $$\mbox A\leq\frac{\int_a^b t^{n-1}\alpha(t)\,dt}{\int_a^b t^{f(n)}\alpha(t)\,dt}.$$  If you want the inequality to hold for all $b>a$, we can take the limit as $b\rightarrow a^+$ and use L'Hopital's Rule:
$$\mbox A\leq\lim_{b\rightarrow a^+}\frac{b^{n-1}\alpha(b)}{b^{f(n)}\alpha(b)}$$
$$\mbox A\leq \lim_{b\rightarrow a^+}b^{n-1-f(n)}=a^{n-1-f(n)}$$
So the best scenario would be to have $A=a^{n-1-f(n)}$, since we'd like to improve on the LHS constant by making it as large as possible.  Now if the inequality holds with this $A$ replacing $a$, then we can rearrange the inequality to read $$\int_a^b\Big(t^{n-1}-a^{n-1-f(n)}t^{f(n)}\Big)\alpha(t)\,dt\geq0.$$  This is supposed to hold for all $[a,b]$ with $a>0$.  This implies that for all $a$, the quantity in the big parentheses is non-negative for $t$ within some $\epsilon$ above $a$, or else we could find an $[a,b]$ that would make the whole integral negative.  So for all $a>0$, for all $t$ slightly above $a$, $$t^{n-1}-a^{n-1-f(n)}t^{f(n)}\geq0$$
$$\Longrightarrow t^{n-1-f(n)}-a^{n-1-f(n)}\geq0$$
$$\Longrightarrow n-1-f(n)\geq0$$
$$\Longrightarrow f(n)\leq n-1$$
At this point we'd like to make $f(n)$ as big as possible$^{\dagger}$, since that makes the exponent in the LHS big.  But then a big $f(n)$ makes for a small $A$!  So there is trade off, and it depends what is more important - having a large $A$ or having a large $f(n)$.  Whatever $f(n)$ is (as long as its $\leq n-1$), the inequality $$a^{n-1-f(n)}\int_a^b t^{f(n)}\alpha(t)\,dt\leq\int_a^b t^{n-1}\alpha(t)\,dt$$ holds for the same reason the original inequality holds:
$$\int_a^b t^{n-1}\alpha(t)\,dt=\int_a^b t^{n-1-f(n)}t^{f(n)}\alpha(t)\,dt\geq a^{n-1-f(n)}\int_a^b t^{f(n)}\alpha(t)\,dt$$
So if there are no mistakes here, the following are each correct, sharp (in a balanced sense) inequalities:
$$a^{2}\int_a^b t^{n-3}\alpha(t)\,dt\leq\int_a^b t^{n-1}\alpha(t)\,dt$$
$$a^{n^2}\int_a^b t^{n-1-n^2}\alpha(t)\,dt\leq\int_a^b t^{n-1}\alpha(t)\,dt$$
$$a^{\ln(n)}\int_a^b t^{n-1-\ln(n)}\alpha(t)\,dt\leq\int_a^b t^{n-1}\alpha(t)\,dt$$
$$a\int_a^b t^{n-2}\alpha(t)\,dt\leq\int_a^b t^{n-1}\alpha(t)\,dt$$
CORRECTION
At the $^{\dagger}$, the desire to make $f(n)$ big really only applies if $a\geq1$, since that would make the integral larger.  If the entire $[a,b]$ is contained within $[0,1]$, then it would be more desirable to make $f(n)$ small.  However, since we are now assuming that $a<1$, that would still make $A$ small (still not desirable).  Again there is trade off, and the above inequalities are still true and (balanced) sharp.  Lastly in the case where $[a,b]$ straddles $1$, we could break the integral into two parts, and the inequalities would still hold.
At the $^{\ddagger}$, the argument that follows only holds for $A=A(a,\alpha,n)$, not $A=A(a,b,\alpha,n)$.  So any conclusions still leave open the possibility for improvement when $A$ depends partially on $b$.
