How to prove that $\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$ for $p >1, x\ge0$ 
Show that for $p>1$ and $x \ge 0$:
$$\dfrac{2}{\pi}\int_{x}^{px}\left(\dfrac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$$

My idea is to use $$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\cdots $$
 A: Substitute $x=e^y$, $p=e^q$ and $t=e^u$.  The problem becomes 
$$\frac{2}{\pi}\int_y^{y+q}e^{-u}\sin^2e^u du\le1-e^{-q}$$
Next,
$$f(u):=e^{-u}\sin^2e^u\le e^{-|u|}$$
If $u>0$, it is because $\sin^2 e^u\le 1$.  If $u<0$ it is because $|\frac{\sin e^u}{e^u}|\le 1$.  
If $y<y+q<0$, or $0<y<y+q$, the integral of $e^{-|u|}$ is clearly less than $1-e^{-q}$.
Now assume $y<0<y+q$:  
$$\int_y^{y+q} e^{-|u|}du = 2-e^y-e^{-y-q}$$
which is maximized at $y=-q/2$ resulting in
$$\int_y^{y+q} e^{-|u|}du = 2-2e^{-q/2}$$
Such that
$$2-2e^{-q/2} < \frac{\pi}{2}(1-e^{-q})$$
given that $p<\dfrac{\pi^2}{(4-\pi)^2}=13.4$.
Now given that
$$\int_{-\infty}^{\infty} e^{-u}\sin^2e^u du =\pi/2$$
we want the tails of the integral to add up to more than $\pi /2p$.  


*

*If $u<0$:
$$e^u-\frac{e^{3u}}{3} \leq f(u) \leq e^u$$
then
$$e^y-\frac{e^{3y}}{9}\leq\int_{-\infty}^y f(u) du \leq e^y$$
so the tail is bounded by $x-x^3/9$ and $x$.

*If $u>0$:
$$f(u)=e^{-u}(1-\cos 2 e^{u})/2$$
The oscillation is both diminishing and speeding up, so 
$$\int_{y+q}^{\infty}e^{-u}\cos 2e^{u} du \leq \int_{y+q}^z e^{-u} du$$
where $e^{y+q}$ and $e^z$ differ by $\pi/2$.  Therefore,
$$\int_{y+q}^{\infty}f(u)du \geq \int_z^{\infty}e^{-u}/2 du = e^{-z}/2 = \frac{1}{2xp+\pi}$$
We now need to show the sum of the two tails is at least $\pi/2p$; we need:
$$ p\left(x-\frac{x^3}{9}+\frac{1}{2xp+\pi}\right)\geq \frac{\pi}{2}$$


*

*If $xp<\frac{9\pi}{16}$ then the third term is at least $\frac{8p}{17\pi}$ which we can take at least $\frac{\pi}{2}$ when $p>13.4$.

*If $xp>\frac{9\pi}{16}$ then the first two terms are at least $\frac{\pi}{2}$.


Thus we have covered all values of $p$, for $1 < p < 13.4$ and $p>\frac{9\pi}{16}$. 
A: This is quite a difficult problem, and I found it very enjoyable.  Here is the solution I found:  
First, we give some simple bounds when $x$ is large, or $px$ is small.  If $x\geq\frac{2}{\pi},$ then by using the bound $|\sin(t)|\leq1$,
we have that
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\int_{x}^{px}\frac{1}{t^{2}}dt=\frac{2}{\pi x}\left(1-\frac{1}{p}\right)\leq1-\frac{1}{p}.
$$
Similarly, if $px\leq\frac{\pi}{2}$, then since $\frac{\text{sin}(t)}{t}\leq1$,
it follows that 
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\left(px-x\right)=\frac{2xp}{\pi}\left(1-\frac{1}{p}\right)\leq\left(1-\frac{1}{p}\right).
$$
Now, assume that $0\leq x\leq\frac{2}{\pi}$, and that $px\geq\frac{\pi}{2}$.
Then notice that 
$$
\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt=1-\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt-\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt
$$
since $\int_{0}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt=\frac{\pi}{2}$.
We will now find a bound on the other two terms. Working over an interval
of length $\pi$, by pulling out a lower bound for $\frac{1}{t^{2}}$,
we have that for any $y$ 
$$
\int_{y}^{y+\pi}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{\left(y+\pi\right)^{2}}\int_{0}^{\pi}\sin^{2}(t)dt\geq\frac{\pi}{2}\int_{y+\pi}^{y+2\pi}\frac{1}{t^{2}}dt,
$$
and so 
$$
\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\int_{px+\pi}^{\infty}\frac{1}{t^{2}}dt=\frac{1}{px+\pi}.
$$
Since the function $\frac{\sin(t)}{t}$ is monotonically decreasing
on the interval $\left[0,\frac{2}{\pi}\right],$ it follows that for
$x\leq\frac{2}{\pi}$ we have 
$$\frac{1}{x}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\int_{0}^{\frac{2}{\pi}}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\cdot\frac{5}{3\pi},$$
and hence 
$$
\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{5x}{3\pi}.
$$ 
Now, notice that since $px\geq\frac{\pi}{2},$ and $p>1$, by plugging them in directly, we have that 
$$
\frac{5\left(xp\right)^{2}}{3\pi}+\frac{2}{3}px+p-\pi>\frac{5\pi}{12}+\frac{\pi}{3}+1-\pi=1-\frac{\pi}{4}>0.
$$ 
Rearranging the above by dividing through by both $(px+\pi)$  and $p$, we obtain the inequality
$$\frac{5}{3\pi}x+\frac{1}{px+\pi}>\frac{1}{p},$$ 
for $px\geq\frac{\pi}{2}$, and $p>1$. It then follows that
$$
\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt+\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{p},
$$
 for $x\leq\frac{2}{\pi},$ and $px\geq\frac{\pi}{2}$, and hence
we have shown that for all $x\geq0$, and all $p>1$, 
$$
\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq1-\frac{1}{p},
$$
as desired. 
