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 $p >1, x\ge0$

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.

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 $\pi /2p$.

1. 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 $x$.

2. 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: $$ p\left(x-\frac{x^3}{9}+\frac{1}{2xp+\pi}\right)\geq \frac{\pi}{2}$$

1. 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 $p>13.4$.
2. 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 $p>\frac{9\pi}{16}$.